# Using Integration by Substitution (by Example)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] okay so you know some basic rules for integration
for example the integral of sine is minus cosine plus the constant but what
about the integral of the double angle of sine for this
we can use the reverse of the chain rule by that i mean make a substitution and
use the composition of functions this process is easier after some practice
hi everyone welcome back to the calculus one explore discover
learn series i’m dave and i’m going to take you through this

00:01
in this video i’m going to cover what is the i’m going to cover the substitution
rule i’m going to work out lots of examples so that you get
better skilled at it and then after lots of examples we’re going to work on some
applications and then stick around to the end where we’re going to discuss some
exercises so let’s get started okay so what is the substitution rule first
okay so we’re going to start off by trying to integrate something
that was that is complicated so for example suppose we want to integrate
this expression over here f of g of x times the derivative of g times dx what
if we want to integrate that now before i go into this too far let me
just say this that in the last video we talked about
anti-differentiation we talked about indefinite integrals

00:02
so that video is something surely you want to watch before this one here
because this one takes the integration up a notch
and we’re going to see how to integrate many more functions than we were able to
do last time but the point is is that you have seen some
integrals before so for example you know what the integral of sine
sine x dx is and you know so we’re going to try to
broaden our skill to develop our skill and be able to integrate many more
different types of functions [Music] so um if we try to e
integrate something that looks like this right here
then our goal is to be able to simplify it and the way that i’m going to try to
simplify it is by making a substitution so i’m looking at this inside function
right here g of x and i’m going to call that a u now
another video that you should watch before this one is the video on

00:03
differentials because once you know what a u is you need to be able to find the
differential so the d u will be the derivative of g times dx
and so that is something that you will need to be able to do also
now the u here is a you know sometimes it’ll feel like a
guess what you’re trying to guess here but the idea is that once you choose the
u and if you’re able to find the d u right there then you can try to take this
complicated in looking integral into something that looks
uh you know something that looks easier to work with and then you want to try to
integrate that so whenever you are trying to do
substitution integration by substitution there’s always
two steps to this process the first is to actually make the substitution
and then the second step is to actually integrate continue
integrating so there’s two places where it might fail
maybe you may not be able to make a substitution or if you are able to make a

00:04
substitution then maybe you wouldn’t substitute into something and
you know then maybe you won’t even be able to integrate this
so we’ll look at all of this um you know and lots and lots of examples ahead
so first off though let’s get started and look at a simple
easy example to make sure that we understand this process here
so let’s look at an example here number one
so we’re going to try to integrate here the integral of 2x
x squared plus 3 to the 30th power now i chose 30th power because
um from what we talked about last time you know last time we did stuff like this
x squared plus 3 and then we could square it so last time we talked about how to
integrate stuff like this and the way we did that

00:05
was by expanding this out right so this is going to be x to the fourth
and then plus a 6 x squared and then plus a 9 and then dx
and then now i can integrate this right here by just
expanding it out uh so by expanding it out and then integrating term by term
so this will be x to the fifth over five and then six x to the third over three
plus nine x and then plus a constant okay so this right here should look
familiar to you by watching the previous episode but
this problem here with a 30th power here we’re not going to be able to take this
approach here because i’m not going to expand this out to the 30th power here
that’s just too much work so we need a different idea
and the different ideas this one right here the
you know substitution how does it work so let’s get rid of all this i hope
you’re familiar with this already if not go back and watch that previous episode

00:06
so we need a new idea here substitution rule so i’m going to try to make a
substitution now as i mentioned over here this rule this this inside function
right here is g of x so think of g of x as a function sitting
inside of a function and that’s we have going on right here
we have this x squared plus 3 that’s a function
and it’s sitting inside a you know function that’s raises something to the
30th power so that’s a function right there and this is a function right here
so this is going to be my u of x this is going to be my my my guess my try
at the u of x and i’m going to try to simplify this in all
x’s which is what this one over here is into this one over here which is just
all u’s f of udu so let’s see if we can do that
of course we first need to know what d u is d u is
take the derivative of x squared so it’s 2x

00:07
take the derivative of 3 is zero and so then we have times dx [Music]
so there’s rdu and u and now we’re going to try to
take all of this right here and go to use only
but actually my first step is i’m going to try to write this in
a more friendly way right this is a 2x and i can write the 2x over here
so this will be integral of u to the 30th and then all this right here is du
now really you don’t need to do this first step because all we did was move
the 2x over here but i just wanted to see i just want you
to see it the first time you look at it this is 30th
and so the point is is that this looks complicated
this looks easier it’s the same thing over here
this looks complicated this down here looks easier
we found a u so that that process complicated easier

00:08
why is this easier because this is just the power rule so this is u to the 31
over 31 plus c so now we have integrated now the catch is though
that this is in use our original problem was in x’s
so i’m going to come back here and substitute in here
so this is x squared plus 3 that’s what my u is
so u to the 31 so u to the 31 over 31 plus c and then the last step would be
to of course say what c is where c is a constant where c is a constant
okay so we integrated this problem right here
this complicated looking expression right here
and we made a substitution to do that and so this is the f of g of x
right here times g prime and i wrote it out here

00:09
so you can see it more clearly this is the f of g of x times g prime times dx
and this is the f of u d u right here okay so
let’s try um you know because when you look at this right here it kind of looks
like you know that it kind of looks cooked up right because
this d u which is 2x just happened to be there in the problem
so it kind of looks cooked up and it looks like maybe you’ll
won’t be able to use this very often so let’s look at another example [Music]
so this time let’s look at um an x times an x squared plus 3 to the 30th
in other words what happens if we don’t have the 2 there
in our problem what can we do so let’s see what changes here
so what happens if we don’t have the 2 there
we just have an x there now how does that change the problem

00:10
all right so i’m still going to be choosing this is my u
right because this is the thing that we don’t want to expand out to the 30th
power we don’t want to expand that out to the
30th power so i’m going to make that this inside function right here be my u
now because i don’t have a 2 there i don’t have a 2 here
so we don’t have that 2 there so now i cannot substitute in my d u here
see my d u has to have a 2 in it right the derivative here is 2x
so i cannot take this x dx into a du anymore
i’m going to have to have that 2 there to get a du
so i’m going to put a 2 back but the price of putting a 2 is i have to divide by
2 out here so if i if i put a 2 there which is what i want
so i wanted 2 there so i put it to there but now to

00:11
keep this equals here i have two and so i have to divide by two so two over two
leads us to a one so another way to think about it is like this
we have an x here so i’ll i’ll multiply it times one
right because we know the the x is 1 times x but for the 1 i’ll put a 2 over 2
and because we wanted 2 with the x i’ll leave the top 2 on the n inside here
and this two on the bottom i’ll pull this out of the integral
so in other words you did get a two here and a one half here
so this will lead us to a one half here this one half comes down
all the way through here and then our final answer we’ll just go
ahead multiply that out 62 here so long story short if they only give us
an x here we can compensate by multiplying and dividing by

00:12
two and then we’ll have the one half coming out of the integral so the one
half will come down all the way and it does
change the answer at the end obviously because it’s a different problem but
but um you know so that’s not much of impediment
of not having a constant two because we can
uh multiply and divide by that constant and we can still integrate
okay so let’s look at one more let’s look at number three though
so number three certainly looks like may not work
and in fact integration by substitution does not always work
you cannot always apply it and solve every problem
but here we go let’s try this so the integral of six x of six x to the third
x squared plus 3 to the 30th now i still don’t want to raise this to

00:13
the 30th power so i’m going to call that a u
and i’m hoping that this will somehow work out i don’t know
so i’m going to say d u is 2x dx once you choose your u this has to be the d u
so it’s u that you’re choosing now i have to be able to go from
all x’s to all u’s if i can’t go to all u’s
then we can’t make the substitution so i’m going to try to take this 6x squared
i need one of those x’s i have three x’s x times x times
x i have one of those x’s i could put here and the 6 i can take it as 3 times 2
so i can pull the 2 and an x and write it with that
what are we going to have left over let’s write that out first
so we’re going to have 3 and then an x squared and then an x squared plus 3
to the 30th and then i’m borrowing an x and a 2. so i’ll have 2x dx

00:14
so that’s we get so far right so the six x to the third
i say that six is three times two and that x to the third is x squared times x
so i have my d u which is what we need we need to have the d u
and i have now u to the 30th but what about the 3x squared
well first of all we can take the 3 out let’s just do that real quick
all right now we wrote it in a really friendly form
so this would be three and then integral
now i have an x squared here what is the x squared
the x squared is actually what u minus 3 so that’s u minus 3 there

00:15
and this will be u to the 30th and then this is all d u so
now here’s an example where i was able to go from all x’s
and choose this is my substitution and i was able to go to all u’s
but now the question is can we integrate
this because if we cannot integrate this then this substitution isn’t going to
work for us so the question is can we integrate this and the answer is yes
[Music] i’ll take this u to the 30th times u that’ll be u to the 31
minus and then 3 times u to the 30th so we will be able to integrate each of
these so this will be u to the sorry 3 times u to the 32 over 32 and then
minus 9 remember that 3 here this 3 is for the whole integral so 3 times the

00:16
minus 3 so minus 9 and then u to the 31 over 31 and then plus c
where c is a constant where c is a constant [Music]
so now the last step is we started with all x’s
so this is in use so let’s go back and put what the u is so here we go we have
3 over 32 and then u u is x squared plus 3 to the 32 power
minus 9 over 31 and then u to the 31 power so here’s what u is to the 31 plus c
that’s 31. okay so there is our final answer there
so this one doesn’t look so cooked up does it you might be surprised that you
could do this with u-substitution so i chose my u according to the fact

00:17
that i don’t want to expand this out i don’t want to do x squared plus 3
to the 30th power i don’t want to expand that out so let’s try to substitute it
away and it worked we got this right here is our final answer here
okay so when i’m looking at this procedure here let’s just
quickly re review how does it work first i’m taking something that looks
complicated and it has to have this exact form here in order for this to work
the inside function right here g of x whatever it is
its derivative has to be sitting right here and if
if all that is working out then we have this u
and this d u and then we’ll be able to substitute
in and then we can try to continue on integrating
all right so now that we’ve seen how does it work
of course we still have many more examples to do we’re going to do lots of

00:18
examples and get good at it but that’s a initial description of how does
it work next let’s see why does it work so here’s our theorem right here
and let’s go see that this does always work
and by does always work i mean the fact that
this integral is equal to this integral so if you find this integral
you can always reduce it under these conditions here
if you use a differentiable function and whose range is in interval i
and f is a continuous function right so we need the g to have
the range that’s in interval i because g the output of g is going to go into f
okay so here’s how this is going to work so i like to call this the reverse of
the chain rule it’s going to follow because of the chain rule [Music]
by the way also another video that you can go back and check out in the

00:19
description below is the full playlist but by the chain rule so by the chain
rule hopefully you practice the chain rule a lot we can say the following if
capital f is an antiderivative of f of our given function f then
so if and that’s the notation that we used in our last video
we used a capital f or an antiderivative but here i’m just spelling it out
if capital f is an antiderivative of f the given function here
right so f is given to us f is a continuous function so if if f capital f is an
anti-derivative then f of g of x this function here obtained by plugging g

00:20
into capital f is an anti-derivative of f of g of x times g prime of x
so this is an antiderivative of this function now why since
so remember to say this is an antiderivative of this
that means the derivative of the anti-derivative must
get back to f here and so that’s what we want to show
here that the derivative of this function
is equal to this one and then we’ll know that this is an anti-derivative
of this function here so let’s go find the derivative of this
and we can you do that using the chain rule so the derivative
of this composition of functions here is the derivative of the outside
function leave the inside function alone times the derivative of the inside

00:21
function here now we’re using the fact here now that this is an
antiderivative of f in other words if i take the derivative of
f capital f here i should get little f and this is the derivative right here so
this right here will be f of g of x [Music] and then times the derivative
and so this told us exactly what we needed the derivative
of this one is equal to that one and that means this is the
anti-derivative of this so now we can go and integrate
and find out what this integral is so make the substitution u equals g of x
now if u is equal to g of x then d u is g prime

00:22
at dx times dx and so what happens if we integrate this left hand side here
what is the integral of f of g of x times g prime of x what is that integral
according to what we just said right here what is this integral
right so i’m integrating this right here but all this is equal to
this right here this is an antiderivative so we’ll have f of g of x
plus a constant right because capital f with g inside of it
that is an anti-derivative of this right here
so this the integral of this right here is equal to the anti-derivative plus the
constant this is the anti-derivative here so now f of g of x here right
so g of x is the u so this is capital f of u plus c now

00:23
this is capital f is an anti-derivative of capital f is an anti-derivative of
f so this is equal to the integral of f of u d u
and that’s what we needed to prove we needed to prove that this integral
was equal to this integral right here and it all followed by the chain rule
right here so if capital f is anti-derivative of f
then this is an anti-derivative of this one and so when we write that out as an
integral statement we get that this integral is equal to
this integral right here so that’s why the substitution rule works
given the assumption that u is a differentiable function
which makes sense because when you go to
try to find the u you first step is then to go find the du
in other words g must be differentiable okay so that’s why it should work and

00:24
how it should work and now let’s take our understanding to a new level
and let’s go and work out lots of examples [Music]
alright here’s our first example here now
now this is example that wasn’t like the first three i showed you where
everything was a polynomial and it was fairly obvious
in terms of what the u substitution should be you know because we didn’t want to
expand something out to the 30th power on this example it may not be as clear
in terms of what the u substitution can be in fact there actually might be
multiple ways of solving this problem with substitution i’m going to take the
approach of one way here and the idea that i’m thinking here

00:25
when i’m trying to solve this problem here i’m looking at that square root there
a theta is pretty constant throughout so i’m not going to let that distract me
the square root of theta is in the argument of cosine and in the
argument of sine squared right so what i’m really looking at here
is that the fact that the sign is squared so since the sign is squared i’m going
to use that to motivate me to make that choice here
so my choice is that i’m going to choose u to be sine of square root of theta
and that’s my choice there that i’m going to try to make and this is a try
and see this may not work we’ll have to figure it out let’s
see what we get what will the d u be if this is my choice what would the d u be
so what’s the derivative of sine it’s cosine
leave the inside part alone now times the derivative of the square root of theta

00:26
which is one over two square roots of theta so this is the du right well no
this is the do you use the derivative times d theta
that’s d u so now let’s see if we can go and integrate this so here we go the
integral of cosine of square root of theta over
square root of theta and then sine square square root of theta d theta
is now when you’re first starting out and you’re starting to get good at this
you want to kind of rewrite things to make it easier to make the substitution
as you get faster you get better and you
don’t need to write all these steps down but
you know first example here i’m gonna i’m gonna write it down
so i’m gonna look for my du somewhere you know i already know where my u is my
u sitting right there it’s just gonna be u squared
what about all this other stuff do i have everything i need

00:27
so i’m going to write this as 1 over sine squared theta square root of theta
and i’m going to take the cosine and the square root and put them over here
and ask the question do i have everything i need is this my du
right here so i took my numerator and my
square root of theta and i just put them over here is this d u not yet
we’re missing a 2 down here aren’t we right so we have everything we need
except for the two so i’m going to put a 2 down here
now at the expense of dividing by 2 so we have to multiply by 2.
so 2 over 2 gives us the 1 so the equation is balanced
so now we can make our substitution this will be 2
and this will be the integral of 1 over u squared
and all of this right here is our d u if we check it closely
this is the d u right here and so now we have taken

00:28
our original integral right here which looks very complicated in theta
down into something that looks very easy to evaluate
this integral in terms of u that’s just one over u squared
we can just use the power rule right it’s just u to the minus two
how do we integrate u to the minus two we’re gonna add one and divide right
so we’re integrating that it’s just gonna be u to the minus one over minus one
right that’s the integral of uh of u to the minus two you just add one and
divide right that’s the power rule we talked about last time so this will
be two times so u to the minus one over minus one
plus a constant and i’ll say over here where c is a constant
okay so now the last step would be to [Music]
go from the u’s come back over here this is the u
and so let’s simplify this here so this will be a minus this will be a minus

00:29
two i’ll put it over here like this so minus two over and then u to the first
power so this will be sine of square root of theta and then plus c
right that’s a minus one so it’s in the denominator
and then use just sine square to theta and now the last step of my
might be to actually um write this in terms of cosecant
some people might like like it to see in terms of cosecant
it does look especially nice like that all right so we found this integral
right here it’s just simply minus 2 cosecant square root of theta
that right there if you go take the derivative of this
you’ll you’ll be able to manipulate the derivative to look like that
all right so there we go there’s our first example there
and this substitution worked because we were able to go
from all thetas to only use and we were able to integrate that

00:30
okay so let’s look at our next example okay so here’s another one where there
may be different choices in which you can choose for your substitution
maybe maybe not i’m going to choose u to be
the inside part here that i don’t want to cube i don’t want to cube all that i
don’t want to multiply all that out so i hope this works 7 minus r to the
fifth over 10. so whenever you write your u down it’s always a try and c
it’s sort of like factoring so what’s our d u derivative of seven is zero
and here we’re going to have a minus five come down so it’s gonna be minus a
half and then now r to the five minus one so r to the fourth here
so is that our du no do you use all of that times dr
and there’s our substitution that we’re gonna try

00:31
let’s try this substitution so here we go r to the fourth
is seven minus r to the fifth over ten times dr so i always ask the question do
i need to manipulate this more before i make my substitution to make my
substitution as easy as possible now this is to the third i forgot so this is
u to the third that’s pretty clear do i see my du everywhere well i have my dr
and i have my r to the fourth i have those but i’m missing a minus one half
so how do i do that so i’m going to integrate here seven minus
r to the fifth over ten all that to the third
this r to the fourth i’m going to put over here
but we need more than r to the fourth dr we also need a minus one half
i need a minus one half here so i’m gonna put a minus one half here

00:32
and that’s going to need parentheses so minus one half
r to the fourth dr all right so i’m going to go ahead and put a minus one
half here at the expense of putting a minus one
half here i need to multiply this out here by minus two
so minus two times minus one half that balances out to give me a one so
these are in fact equal to each other this is preserved right here
all right so what do we end up with here minus two integral of this is
all my u to the third and all this right here we finally got as our d u
that’s our d u right there so we went for something that looks complicated
to something that looks very easy to integrate
this is just u to the fourth over four plus c where c is a constant

00:33
so this will be minus one half and now what was our u
our u was all this seven minus so seven minus
r to the fifth over ten so u to the four so all this
to the fourth plus our constant there and there we go so this is our u right
here just substituting in the u and we’re done
okay let’s go on to the next one now oh no this one looks interesting
what are we gonna do on this one right here now
sometimes when you’re integrating you may need to manipulate
the problem before you start and guess and rush on your u-substitution and this
is one of those problems i don’t want to take the whole thing right here is my u
if i do that i’m not going to get anything particularly nice

00:34
so here’s what i mean by that if i choose the whole inside part here is my u
what will my du be well i’m going to have to start doing the quotient rule
and i’m going to get lots of stuff and that stuff ain’t going to be
anywhere in my problem so this is not my first choice
it might be my first thought but it’s not my first choice
i’m going to manipulate this a little bit first before choosing a u so
how do we do that so i’m looking at this
x to the fifth here now when we take the square root of x to the fifth we’re
going to get some left over stuff right so let’s let’s write it like this x
minus one and let’s write it as x to the fourth with an x
right so the square root of x to the fourth that’s just
x squared right so this is just one over x squared and then x minus one over x

00:35
right so this x to the fourth i can take the square root of that part
it’s just x squared so i got x minus 1 over x left
now this makes it look more complicated but it’s actually going to be better for
us we’re going to be able to figure out what the u is by looking at it like this
in fact i want to actually write it one more time
this will be 1 over x squared in fact i’m going to write it out like this
what’s the x over x that’s 1 and then the minus 1 over x
and now i’m going to write it as x squared here
and now when you write it like that now it’s very suggestive
in terms of what the u should be the u should be this input
inside part here so why is that because what’s the derivative of this
what’s the derivative of minus 1 over x right so that’s just like that and so

00:36
the derivative is negative times negative this x is minus 2 is 1 over x squared
right so the derivative is very nice it’s sitting right there
all right so this is my u that i’m going to choose so my d u is 1 over x squared
dx and that’s sitting right here so after some algebraic manipulation to
making it look like this then it becomes much more clear
in terms of what the u substitution should be so this will be the integral of
so this is all u to the one half and that’s all d u right there
now we went from something that looks complicated to something that looks very
easy to integrate this is u to the one half plus one [Music] plus c where c
is a constant all right very good so now we can take all of this right here

00:37
and change back to the u to the x’s so instead of dividing by three halves
i’m going to multiply by two thirds and this u right here is one minus one
over x to the three halves plus our constant
and there we go there’s our solution right there
i manipulated and i’m thinking about that x to the fifth that’s larger than
the square root so there’s going to be some part that we can pull out of there
and then now manipulating this a little bit i can try to figure out what the u
should be here and then once we get it to use all u’s notice it’s
all x’s and then all u’s should never mix them
now integrate with of with with respect to u
and then come back to the x’s we’re done all right so let’s look at another one
ah here we go so we know what the integral of

00:38
e to the u is right the derivative of e to the u
is e to the u so the anti-derivative of e to the u is e to the u plus a constant
so i’m knowing that and that is my reason for making the substitution here
u is equal to the x exponent sine squared theta
and i’m hoping my d u will be sitting right here the sine 2 theta dx
let’s check it out and see so my d u will be
so now here we got to use the chain rule here so
2 sine theta times the derivative of sine theta which is cosine theta
is that my d u no we need a d theta here there we go there’s our d u
now actually it looks like it may not work because we have sine two theta

00:39
but our d u needs to be all of this now in the previous example we worked
out some algebra before we made our substitution in this case we need to do some
trigonometry remember a trigonometric identity
sine two theta is actually equal to two times sine theta times cosine theta so
this is the same thing as two sine theta d theta
right so all this right here is equal to that and that’s just a trig identity
so using this is my du now it looks like we have some hope
so here we go the integral of e to the sine squared theta times sine two theta d
theta will be equal to the integral of e to the u and all this is my du right
here and that is easy to integrate the integral of e to the u

00:40
is e to the u plus a constant and now we can come back with what our u is
so this will be e to the sine squared of theta plus a constant
where c is a constant so something that looked originally pretty complicated
came down to just integrating this by making a substitution
and by using a trig identity to match the du all right there we go
let’s look at our next example okay now in order to use
u-substitution successfully you often have to have a wide variety of
integration formulas on hand because the point is is that you have to substitute
to make it simpler but you still have to continue integrating

00:41
so you still need to know a good number of formulas so last time we talked about
this formula here 1 over the absolute value of x square root
x squared minus 1 and then dx and this is secant inverse of x
and when i’m when i’m looking at this problem up here
i’m looking at that square root and then i see a power of x minus one
so that really reminds me of this formula right here
that we that we covered last time so i’m going to try to make a substitution on
this problem thinking of hope that i’m hoping i can reduce it to this
because if i can if i can get all this i know it’s secant inverse
so i’m looking at that x to the fourth but what i want is

00:42
something to the squared so i’m going to say um let u be equal to x squared
and so that would give me a u squared minus 1 there
all right but what’s the consequences of that what’s d u d u is 2x dx right
so here we go let’s try to integrate the integral of
one over x and then x to the fourth minus one
dx how are we going to write this out here
let me just write it over here this dx i really need a 2x dx i need a 2x dx
so i’m going to multiply top and bottom by 2
and an x so we can write it like this i guess
you can write it like this one x and then x to the fourth
minus one and i’m gonna say two x over two x dx

00:43
so i’m just thinking about that as 2x over 2x now
you don’t really need to write that step down because that’s just what you’re
thinking with this 2x i need to leave it with my dx
so this 2x i need to leave it here now we can pull the 2
out of the integral but we’re integrating with respect to x
i cannot pull that x out so i can get a one-half here integral
so this x is going to have to go with this x and so i’m going to get 1 over
x squared and then x to the fourth minus 1
and then i still have my 2x dx over here so this would be a good step to write
next i wrote that extra step out just so you can see it
i need to put a 2x in here at the expense of putting a 2x
i divide by a 2 and i divide by an x but i have to leave that inside and so it
combines with that x already there now i’m going to try to

00:44
go to all u’s i wrote it in a very friendly form
so this would be one half integral 1 over so what is an x squared x squared is
just a u so this is u and this is so what’s x to the fourth
so this you know square both sides so x to the fourth will be u squared so this
will be u squared minus one and this was the whole idea of trying to
make this substitution is so i can get a u squared minus 1
there and then this is d u sitting right here that’s our d u
so this looks very similar to our secant inverse formula but with one difference
that doesn’t have an absolute value around it but
u is equal to something that’s always positive or zero
so we can put absolute value there because of what u is u is just square
root of i mean u is just x squared so now we have our secant inverse formula so

00:45
this would be one half secant inverse of u plus a constant
and then final step what is that u it’s an x squared so where c
is a constant and now last step come back to the x’s
so one half secant inverse the u is an x squared so there’s our problem there
our first step here was to put a 2x over here divide by 2 and divide by an x
and so then we got it to work out here so that’s a nice sweet problem there

00:46
okay next example okay here we go what should we try to do
so on this problem here i’m looking at um that cosine inverse is my exponent
i know how to integrate e to the u so i’m going to try this and just see
i’m going to try that to be my u now what’s our d
u we need to remember the derivative of cosine inverse
so it’s a cofunction so i remember it’s negative
and i’m going to get square root of 1 minus x squared here
so this is our d u right well almost d u is all this the derivative times dx so
there’s rdu and we’re hoping this works let’s try and see if we can go from all
x’s to all u’s so we have this right here and that’s in

00:47
so we’re going to have to multiply by a -1 so e to that over square root
i need to make that bigger anyways i’m going to put a minus and then i’m
going to move that over there so i want to make it look a little bit more
friendly negative and then e to the cosine inverse x and then minus 1
over square root and then dx maybe put parentheses if you want
that’s cosine inverse all right i think that looks a little
more friendly so i just took this right here and put it
and put it over here we have a minus minus all right so now i have my d u
sitting right there so this will be negative
integral of e and that’s our u up there and all this is our du

00:48
so we went from all x’s to all u’s by making the substitution
now we can integrate so this would be minus e to the u plus c
and now we can come back in with what the u is so this will be minus e to the u
plus c so just substitute in what the u is so where c is a constant
and there we go there’s our problem right there
integrate that choose that as the u that’s the d u try to find the d u at all as
friendly as possible make the substitution integrate substitute back

00:49
all right next one all right so what should the u be
okay so the u i’m going to try is sine inverse
and it’s sitting right there now if that’s my
u then what are we going to get for all of this right here
1 minus y squared how’s that going to work
so i’m just tentatively using this right here as my u
what would the d u be so remember the derivative of arc sine will be 1 over
square root of 1 minus y squared so that’s the derivative of sine inverse
right there and then we have a differential so d y
so it looks like that’s our d u is sitting right there for us
so this right here we’ll give this a try
the integral of 1 over sine inverse of y

00:50
times square root of one minus y squared d y
this will be the integral of this will be
one over u and then all this right here is the d u
so this will be natural log of the absolute value of u plus c
so that’s u to the minus 1 [Music] and so that will be natural log of
absolute value of u now we come back here u was sine inverse of y plus c and so
that’s it that one looked complicated but actually if you choose the right u now
another choice of u might have been say one minus square one minus y squared
but if that had been your choice for you what would you have done with sine

00:51
inverse of y okay so let’s look at the next example
all right so this one right here to me when i see something that looks
like this i’m thinking there might be multiple choices for
our u substitution but when i’m looking at this
i’m going to try to focus my attention on cosine to the third
so cosine looks like it’s a function inside of a
function that’s raising something to the third so that’s going to be my guess
i’m going to try this is my u to be what’s inside that third power so i’m
going to try cosine of square root of theta
and see if that works so my d u will be so the derivative of cosine is minus
sine and then leave the angle alone now times the derivative of the angle
square root of theta which is one over two square roots of theta

00:52
and then d theta so that’s my u and so that has to be my d du
and so let’s see if we can make this work so the original integral here is
sine of square root of theta over square root of theta and then
cosine cubed square theta and then d theta and so let’s try to make this look a
little bit more friendly before we do our substitution
so first off i notice that i’m going to need a negative because of this right
here so i’m going to need a negative with all
this stuff here so i’m going to pull a negative in here and multiply by a
negative so i’m going to say this is negative and then integral
and then let’s see here we’re going to have 1 over
cosine to the 3 halves square root of theta power

00:53
so that’s a cubed but then i have square root on that
so it’s going to be three halves and what do we got left over we still have a
sine square to theta and we have a square root of theta here
and so it looks like we’re missing a 2 down here so i’m going to put a 2.
in fact it looks like we’re missing a minus 2 here so i’m going to say minus 2
here and then i’m going to put a minus 2 out here
so i’m dividing by minus 2 so i multiply by minus 2
and then we have d theta so let’s double check everything
our sine squared of theta is still here we have the square root of theta right
here we still have that we have square root of cosine to the third good
we left the angle alone now we need a minus two coming in here a minus and a two
so i put it here and then i balance and then last step is maybe to put
parentheses around this right here all right so now it looks plainly

00:54
obvious this is minus two this is integral this is one over
u to the three halves so i’m just gonna say u to the minus three halves here
because it’s in the denominator and then all of this
is our d u right we got minus sign and then one over
so that’s all our d u right there all right so after all of that work there
choosing this is our u because of this power down here
i get this nice power here and then this is our d u luckily all that stuff
is there all right so here we go minus 2 now we’re going to integrate
so let’s integrate here so u to the minus three halves plus one will be
minus one half divide by minus one half plus c so we gonna add one to the
exponent and then divide and then we’re gonna say here equals

00:55
so this is dividing by minus one half which is the same thing as multiplying by
minus -2 so i have a minus 2 times -2 which is just 4 and this is a negative
exponent here so it’s going to be square root and then what is the u
it’s cosine to the square root so i’m going to put cosine square root of theta
in here and then plus c where c is a constant
so i put the u back in and i simplified it all
minus two dividing by minus one half is just
four and that’s a negative exponent so i wrote it in the denominator here
all right so that looks good so we were able to
calculate integrate this whole mess here it just happened to work out for us

00:56
didn’t it all right next example ah something different here so let’s
read this here it says the um slope of each point x x y
on the graph is given by now it’s saying
the slope at each point is given by that means we’re given the derivative
so let’s write that down before we read the rest of the problem
the y over dx is given to us the slope at each point
so that tells me this is the derivative all right and the graph passes through
the point so we’ll have to keep that in mind but
what we’re trying to find is capital f here find
f find find the function and for which this is the derivative
so we’re going to have to integrate so f of x is the integral of x to the

00:57
x squared minus one to the one third dx so this is the derivative and we’re
trying to find the original function f so if you integrate the derivative you
should get back to the original function f now we try to integrate this
we don’t have a formula let’s use u substitution
so u is the x squared minus one and so what would d u be d will be 2x dx right
so do we have a 2x dx here well we can compensate for that missing 2 there so
let’s write it out as one half integral of x squared minus one to the one third
and then so we put a one half so now i have a two and an x dx
so that step may be optional by now but you know if you’re if you’re still
getting the hang of it that might be your next step is to

00:58
write it out make it make it look friendly maybe you make less mistakes that way
in any case we have one half the integral of this is our u to the one third
and this is our d u here okay so now we can integrate this is one half
u to the one third plus one so that’ll be four thirds
divided by four-thirds plus c so add one and divide
so dividing by four-thirds is the same thing as multiplying by
three-fourths so this will be three-eighths
and then this u here is x-squared minus one so x-squared minus one to the
four-thirds plus c there we go so we found the function f of x
it’s this right here uh actually c is a constant right

00:59
where let’s write that down where c is a constant
where c is a constant now they tell us the graph passes through that point and
the reason why they tell us that is so we can find the c when we plug in
3 we should get out one what happens when we plug in
three here so when we plug in three so let’s do that over here
so f of three we plug in three here we’re gonna get three eighths
times so we’re gonna have three so it’s gonna be nine
minus one so that’s eight when we do the cube root of eight that’s two
two to the fourth that’s sixteen and then plus c must be equal to one
when it’s the original problem says when you plug in three you have to get out
one so i plugged in three and i got out of one and so c has to be here um

01:00
so that’s a six so so minus five so c is minus 5.
so now we have our function let’s go down here and write it
so f of x which we found by integrating all that
is this and then we found our c to be minus five so we have three eighths
and then x squared minus two x squared minus one
four thirds and then minus five as our c so there’s our function there
so for that function right there we can you know
come back to our original problem and make sure it makes sense
knowing that this is our derivative right here
and it passes through three one here’s our function right here
if you were to go graph this function right here it passes through 3
1 and if you were to take the derivative of this
you would get that x times x squared minus one one third

01:01
all right so next problem okay this is a fun problem here i like
this problem here all right so how do we evaluate this integral here
i’m going to notice that what we did before was that sine 2 sine cosine
x is equal to sine 2x remember we we used that trig identity
on an earlier problem so our given integral here is the integral of sine 2x
so this is just sine 2x dx so i’m really if you want to integrate this you can

01:02
integrate this so what i’m going to do is use my u as a 2x here
so u is 2x the inside part right here the angle of the sine so du will be 2
dx so that’s the substitution i’m going to try there
now i don’t have a 2dx because that 2 is with the angle
so i need to put a 2 here so i’m going to say one half integral of sine 2x
times 2 times dx so this 2 is going here with the dx i
compensate by dividing by 2 and i still have my sine 2x here
so this will be one-half integral of sine of u and that’s all my d u
now we know the integral of sine right minus cosine so i’m going to have

01:03
minus one half cosine u plus c and so this will be minus one half
cosine and then the u is the 2x so 2x plus c and i’ll just put over here
where c is a constant okay so there we go i used a trig identity
and to get it to something simpler and then i used my 2x for my u
and that looks good to me all right now what if you didn’t think to

01:04
and choose a different u maybe you say try u is equal to sine x
and i’m looking at that problem right there the integral 2 sine x cosine x dx
what if i said ah i don’t i don’t remember that trig identity
what if i use this as my u so what would d u be
the d u will be right the derivative of sine is cosine
dx and we have a cosine dx sitting right there
so let’s try to integrate this the integral of 2 sine x cosine x dx
is i can pull that 2 out so 2 integral of sine sine x dx cosine dx
so just pulling the 2 out right so this will be 2 integral of
the sine x is the u and the cosine x dx that’s the du so this would be udu

01:05
[Music] all right can we integrate this yes this will be 2
and then this will be u and then we add 1 and divide and then plus c
there we go now actually the twos cancel and what was the u the u was sign so
this would be sine squared and then plus a constant where c is a constant
where c is a constant so this u right here is just sine squared
and the twos cancel which we have plus c there we go
so when we use the trig identity we got this is our answer cosine 2x
and here we have sine this is our answer so the question is which answer is
correct [Music] which answer is correct now the key is

01:06
that there are both correct and the reason why is because
you can use a trig identity to go from one to the other
and you have to realize the fact that the constant is arbitrary
and so can look like different things for example um
we know that sine squared plus cosine squared
is one so if i when i’m looking at the sine squared here
i can substitute in here one minus cosine squared plus c well 1 plus c
is a constant so i could actually write this answer like this also
where c is an arbitrary constant in fact don’t try this as your u
go back to the original problem and try a

01:07
different u so hey let’s do that for fun come on let’s do that
what if our u is cosine x have we tried that our du will be
what’s the derivative of cosine it’s minus sine
so if i’m going to integrate 2 sine x cosine x
i need a minus sine x minus sine dx i don’t have a minus sign dx but we can
get one i’m going to pull the two out and put a minus in front
so i still have my cosine but for this sign i’m going to put it over here with
a minus sign wait what are those two things equal what’s minus 2 times the minus
positive 2. and we still have sine times the cosine

01:08
yeah these are equal to each other but now i have my d u don’t i
that’s my d u right there and that’s my u so this would be minus 2 integral of u
and then d u so integrating this this will be minus
two and this will be what u squared over two plus a constant twos cancel
and what was our u again cosine squared [Music] all right so that’s the answer
also so depending upon which substitution you used
all correct by using trig identities you can move from one of the
so that’s why i like that problem is there’s a variety of ways of natural

01:09
ways to do this problem and it may seem like your answers are
all different but they’re not you can use the trig identities
to go from one answer to another answer alright so let’s look at um you know
what is next all right let’s do some applications
all right our first application is going to be solving
a an initial value problem now i call this an application because
this is a differential equation and solving differential equations often
comes up in physics you know a variety of scientific fields
but quite often in physics and so we’re going to try to solve this problem here
um and let’s see how to do this you know we can do this using integration here

01:10
[Music] so let’s let’s get started on it um now in order to solve an initial
value problem is it gives you a second derivative and our
goal is to find the original function and we have to use these boundary
conditions these initial conditions to determine the constants of
integration that we’re about to get so let’s get started so before we find
the original function we’re going to first find the first
derivative and we’re going to do that by integrating so we’re going to integrate
4 secant squared of 2x and then tangent 2x now we solved some initial value
problems before in the last video however in the last video we weren’t
able to integrate such expressions such functions like this
because we weren’t very aware of the substitution rule yet

01:11
i’m going to make a substitution i’m going to say u is equal to
which one do you think we should use now secant
has got the power but actually i’m remembering
that the derivative of tangent is the secant squared
so my choice should be tangent of 2x all right so now there’s our choice and
the reason why is because the derivative of tangent
the secant squared and then i’m going to have a 2 out so i’m going to have 2
secant squared 2x times the derivative of 2x which is the 2 and then the dx
[Music] okay so this is the substitution i’m going to try
tangent 2x and then this is my du now does it look like we have our du
yes but our constant needs to be changed here
so i’m going to write this as integral of this is my u right here
and this is my 2 times d u here so i’m getting

01:12
2 u d u or if you want to write it out with a u out here that’s fine
as well with the 2 out here as well right because that that’s only got a 2
in it so think of that 4 is 2 times a 2 and one of those twos went with our du
and the other two just stayed around in any case
this is going to be if we come back to if we integrate this
this will be 2 and then u squared over 2 plus a constant and then coming back
here with the 2 this will be tangent squared of 2x plus our constant
now what is our constant so we’re given this initial condition the
derivative at zero is four so here’s the derivative i need to plug in
0 into my derivative so if i look at my derivative i’m going to plug in 0

01:13
and my derivative here and i should get out of 4. so um y prime
at zero is so plug in zero and we get out c
and we should get out of four all right so c is four
and now we have our derivative our derivative is equal to tangent squared two x
plus our constant four so there’s our derivative
tangent squared two x plus four because the c is 4
and we found the c4 by plugging in 0 and getting out a4
all right so now in order to find our original function
we need to integrate the derivative so here we go

01:14
y is equal to the integral of tangent squared two x plus a four okay here we go
now how do we integrate this tangent squared um we don’t have a formula
for integrating tangent squared at the moment so we’re gonna have to rely upon
a trig identity let’s remember that tangent squared plus one is secant squared
right remember that trig identity wait you don’t remember that trick identity
well maybe you remember this one sine squared x plus cosine squared x is one
what if you divide everything by cosine squared so tangent squared plus one
equals secant squared so if you divide everything by cosine
you get tangent squared plus one equals cosecant squared

01:15
so this will help us with the tangent squared
here so when i’m trying to integrate this
i’m going to use this trig identity that tangent
of something is secant minus one let’s just move that over equals secant
minus one right so here we go tangent squared so
this is going to be the integral of instead of tangent squared i’m gonna
write secant squared of the angle minus one and then we still have a plus four
[Music] all right so let’s just write that real quick that’s secant
squared two x um plus three [Music] right putting together that four and three
[Music] now we do have an integral formula for secant squared right
because the derivative of tangent is secant squared
and we can integrate this term by term right
so when we integrate the first one here we’re going to need

01:16
a u substitution though so let’s write it out secant squared 2x
dx and then integrating the three right this one doesn’t need a
substitution but this one does because i have a 2x there
so i’m going to say u is 2x now we already used the u in this problem so
i’m going to actually just make it easier to someone to read
say v is 2x so dv is 2 dx so this will be a v right here and i need a 2 dx
so i’m going to multiply and divide by 2 so one half integral of secant squared
2x and then times a 2 dx and then this one is still 3dx
all right so now we have our dv right here it’s sitting right here

01:17
right that’s our dv right here so this will be
one half integral of secant squared v dv and then this integral right here
and now we have formulas for everything we have a formula for secant squared and
we have a formula for three so this will be
one half and this will be tangent of v and this one right here will be three
3x now for this v because i’m running out of room right here i’m just going to
say the v is the 2x and put it in here and then then plus c all right so it’s
one half tangent of v or 2x right that’s the
integral formula the integral of secant squared is tangent so tangent of v and
tangent of 2x and then integrating just the three right

01:18
and then plus c now we have some information to find the c
right because this is equal to the original function this is the original
function that we just found we traced back the equal signs all the
way this is the original function y here and if we plug in zero into the y
we should get out minus one so if we plug in zero here everywhere right so
let’s do that over here so y of zero we plug in zero everywhere what’s
tangent of zero that’s zero and then we plug in zero here
so out of all that we find c equals minus one
so we plug in zero we have to get out minus one but when we plug in zero we
get out c so c is minus one so now we have our function y
so i’ll just write down here y is one half tangent two x plus three x

01:19
minus one there’s our function there and if we go and take the second derivative
of this function we get that one up there if we plug in zero into this
function we should get out minus one and if we plug in zero into the derivative
we get out four and there’s the work there
we started off with this u substitution here
we integrated we found the constant we found the first derivative
we integrated the first derivative to get to the original function
that took a little bit of work because it was a tangent squared we don’t have a
formula for tangent squared we do have a formula for secant squared
so we integrated we found our constant and we got our function there
excellent now next application all right so suppose it’s known

01:20
that in a certain country the life expectancy
at birth of a female is changing at a rate and that so they give us that this is
the rate and they did a lot of statistical analysis
or or whatever techniques they were able to come across
to come up with this rate of change here this derivative
here t is measured in years and we started this
in where they started this in 1900 and they want us to find g
so that we can start making some predictions
once you know g then you can plug in values now we can integrate to find the g
because we because we’re given a derivative if we
integrate the derivative we’re going to find the original function g
now they give us some initial conditions so we can find the constant of
integration and then they ask us a question in other

01:21
words they want us to make a prediction use g to make a prediction
so let’s get started on this so g is going to be g of t
is going to be integrating the derivative and so this is going to be integrating
the 5.45218 all over 1 plus 1.09 t and then all that is to the
0.9 power and then dt so to integrate all of this right here
we’re going to make a substitution and we got all this right here with the t in
it to this power here so we’re going to make a substitution
i’m going to say u is equal to this part right here

01:22
so u is equal to 1 plus 1.09 t that’s this inside part here so d u will be
1.09 dt so i need a 1.09 in front of my dt which
we don’t have we have this number here instead in front of it so we can
use the substitution here and we can pull out this number here
and put in this number here but we need to divide by it also so 5.45218
divided by the number we need 1.09 integral of 1 over 1 plus 1.09 t to the
0.9 and then now since i divided by it i’m going to multiply by it and then i
have a dt okay so hopefully you can tell that these two are the same here
we just pulled that constant out i divided by this 1.09 and i multiplied by 1.09

01:23
and then this is still i just left all this alone
now this makes it very friendly to to use our substitution with it
because this is our d u and our d is sitting right here
and this is one over u so here we go we have the
equals two this number here five point five two one eight
1.09 and then this will be the integral of 1 over u to the 0.9
and then that’s d u right there all right so now our integral looks much
much easier to do in fact maybe you didn’t even write it
like that maybe you wrote it like u and since this is in the denominator
negative 0.9 and then do you so maybe just wrote it like that
but whatever we’re going to now add one and divide so now we get 5.45218

01:24
over 1.09 and then now we’ll integrate this we’re going to get u to the
negative 0.9 plus 1 so we’re going to get 0.1 over 0.1 plus our constant here
and this u we know it’s equal to that 1.09 times t i’ll go put that in
and so we’re going to get 5.45218 1.09 and then times 0.1 down here
and then we’re going to get u which is 1.1.09 t to the 0.1 power
in other words just plugging in the u and then plus c
just plug in the u so we got this crazy number here
and if we go crunch that number out we’re going to get 50.02

01:25
one plus 1.09 t 0.1 plus c and this is going to be our g of t that
we’re looking for so g of t we integrate the derivative so
we integrate the derivative using a u substitution showing
every step for you possible this is our g of t right here let’s just write that
down here g of t here now to find this constant here
they give us some information over here so when at the beginning 1900
it’s 50.02 which is what we already have here so g of zero when we plug that in
what happens we plug in zero here right when it’s 1900 that’s
your zero so when we have zero here that’s going to be one to this power so
that’s all one so we’re going to get 50.02
plus c and they tell us over here that that should be equal to

01:26
50.02 so c is in fact zero so this is our function right here and c is zero
so the last step is to now answer the question
what is the life expectancy at birth of the female born in the year of 2000
in in this country here so in 2000 that’ll be in 100 years so now we’re
going to go use our g of t which we just found is the 50.02 times um well
here it is so we’re going to plug in 100 into here so 100
we’re going to plug in 100 here and let’s see so let’s do that over here [Music]

01:27
so let’s plug in 100 and we’re going to get the 50.02 times
okay so i ran out of room here um let’s just plug it in here so g
of 0 of 100 so we’re looking at the 2000 now so g so we’re plugging in 100
100 years later it’s 50.02 and then now we’re plugging in 100 here
so we’re going to get 110 to the 0.1 and then our constant was 0. so we need
to approximate this number right here and we’re going to get 80.04 years
so 100 years later whatever they were doing was pretty good because their life
expectancy greatly increased okay so there’s that problem there
we are given the derivative we integrated to find the original
we used some initial conditions to find the constant of

01:28
integration and then at the end we used our
function to make a prediction of what’s going to happen 100 years later
okay so next thing um this is going to be a really nice
application for us here i’m not going to assume you know anything
about special relativity but because this is just a calculus video
so according to einstein’s special theory relativity the mass of a particle
is given by [Music] um so this is the mass the the m
is going to be given by this expression here m m naught is the rest mass of the
particle v is the velocity of the particle and c is the speed of line
so we’re assuming that here c is a constant and we’re assuming the rest mass is
known and we’re assuming that the velocity is a function of t

01:29
if we were to look at say the limit right here as v goes to infinity
the rest of m so what should this limit be right so we’re going to infinity
well um actually let’s go to c not infinity
sorry the reason why this is interesting is because what’s 1 minus 1
right so as v approaches c then we’re going to get c squared over c
squared which is one we get one minus one we get zero so
what’s happening when the velocity of the particle
approaches the speed of light so we’re gonna get a fixed number m naught over
zero this is going to zero so we know from from previous video

01:30
this is going to be some infinity right here and so now the question is
is it positive or negative infinity well
it’s the square root and that’s positive
so it’s going to go to positive infinity here in other words as the velocity of
the particle approaches the speed of light the mass right this is the
mass the whole thing is the mass is going to approach infinity so your
mass is going to get larger and larger and larger as the
velocity approaches the speed of light in any case let’s continue on with this
example here suppose that a particle starts from rest
at time zero and moves along a straight line under the action of a
constant force so according to newton’s second law of motion
the equation of motion is given to us as f the force is equal to the rest mass

01:31
times the derivative of the velocity so we’re thinking acceleration there but
we’re dividing by the square root of 1 minus the velocity squared over square
square root of c so that’s second law of motion there and our goal as a calculus
student here someone learning calculus is to find the velocity
and position functions of the particle so we’re given derivatives
and we want to find the integrals of those derivatives
so let’s do that oh one more question what happens once we complete all that
we can ask the question what happens to the velocity as the
of the particle as the time goes by so we’ll be interested in that question too
all right so anyways from the newton’s second law there
we’re getting this a force and we want to uh you know solve for the
derivative there so we get f equals f over the initial mass

01:32
in other words just divide both sides by the initial mass
so we can take the derivative so the derivative here
will be once we solve for the derivative we can integrate both sides
so if we integrate the derivative then we just get the
v over square root of one minus v squared over c squared
and so we integrate the left side and we integrate the right side
of course we’re integrating with respect to t the time the variable
now if we integrate that because the f over the
uh initial rest mass all that’s constant remember this is a constant force
so those are constants when you integrate with respect to t
so we’re going to get the constant times t plus another constant k
and k can be uh case of constant and we’re going to find k by knowing that

01:33
the velocity is 0 when we start it out we started out rest so we plug in 0
into the velocity we’re going to get out the k is 0. so
there we go we have the left-hand side and we have the right-hand side because
our constant is zero and now we have to go solve that for v
so because on the left-hand side we have v over
and then we got an expression with v in it right so we need to solve that for v
so multiplying it out and moving all the v’s back to one side we can solve for v
right there and we get that expression right there the velocity
is c times f times t over that expression there
and so that will give us the velocity and now we can go find the position
so the position function will integrate the
velocity will integrate the derivative and so we’re going to integrate that

01:34
velocity function that we just found and we’re going to make a substitution
you know this is a substitution rule video
so we’re going to have that square root in the denominator
and we’re going to choose that as our u and then our d u
if we take the you know differential there the m naught squared c squared that
right there is constant so this part right here is constant so
the derivative of that is 0 and then we’re going to have a f squared
is constant so there’ll be f squared and then we take the derivative
so we’re just doing 2t dt and with the constant f squared in there
also so there’s our du and now we substitute it back in
and we pull out the constants and so we needed d u up here right so we need a

01:35
2 f squared and a t and we have the f and the t already
so we need to put a 2 so we’re going to divide by 2
and then we’re going to divide by the f squared there so we get the
1 over u square root and the d u and then we integrate and then we come
back and solve and plug in what the u is all right so we’ve seen all of that how
to do all that and now we get the position function there right
so we did enough examples in this video hopefully
that we were able to follow this right here and we got
our position function now so now we can try to find the constant of integration
to fully uniquely determine that position function
when we substitute in zero our position is zero
and so that will allow us to solve for the constant
and then right so we plug in zero right so there’s a zero there

01:36
and then we got a square root on that so that’ll give us the m naught
and then the square root of the c squared that gives us a c we already
have another c so that’ll be c squared and so on so there’s the c
so now once we found the c and we have our c here and so now we
have our uniquely determined position function here
so now we can ask the question that they asked us at the beginning
what happens to the velocity as time continues on
as time continues on the long term behavior of the velocity
is you know by looking at what the velocity function here is this was the
velocity function right here and if we look at as t goes to zero uh
as sorry as t goes to infinity well we can uh you know
just take this limit right here and to find this limit it’s probably
just easiest to divide by the highest power
we saw how to do that in the episode on horizontal

01:37
isotopes and vertical isotopes finding limits [Music]
involving infinity but anyways dividing by the highest power here
we’re going to get that uh this expression here
now as t goes to infinity this part right here is going to go to zero
and so we’re going to get square root of f squared
those are going to cancel the f’s are going to cancel the only thing remaining
is the speed of light so the velocity approaches the speed of
light as time goes on long-term behavior there all right let’s
look at some exercises now [Music] okay i got lots of exercises here
and the idea is to not just theoretically know how use substitution works
and not just to work out a couple simple examples and say oh i got the right

01:38
answer but you really want to get really skilled at this
because you’re going to see integration by substitution
throughout calculus 2 and you’re going to see it throughout calculus 3
and you’re going to see it throughout differential equations
and you’re going to see integration by substitution
really needs to be something that you can do and accomplish
very quickly in fact when you get to calculus 2 you’re going to study a whole
a whole chapter on techniques of integration and
throughout it is going to be integration by substitution it’s just
you’re just going to keep getting better and better at that so
practice these 14 problems here and then here we have some more
so these are going to give you some words and then you have to go integrate
and do substitution and probably find some integrations of constants
using some specific values so here’s some more to practice on

01:39
and then we have seven to get even uh more challenging problems for you
right and so after you do these you get get even better
now on these problems here are actually formulas that are given to you so one of
the things you should be able to do is say
if this is a formula ah let me check it someone’s claiming
it’s a formula but how do i know it’s actually a formula
how can we check for example on problem number eight how can we check number one
is actually a true formula so what you can do is you can take the
derivative of the anti-derivative and get back the original problem and
that will help you check if that problem is correct so for
example number three right here how do i know that this is the correct
integral formula we go take the derivative of the antiderivative

01:40
all right so on on same thing on number nine okay on 10 and 11 here
uh well in 10 we’re going to practice solving some initial value problems
so either a first second or third derivative is given to you
and you have to go find the original function and to determine the constants
of integration using the initial values of the functions now number 11
we’re given some f and g’s and we need to put together that information to find
those integrals all right so i want to say thank you for watching
if you like this video make sure and hit that subscribe and like button
in the next video we’re going to talk about um
definite integrals and area limits of riemann sums uh it’s going to be a blast
it’s one of the funnest videos i love it and i look forward to seeing you then

01:41
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