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# Trig and u substitution together (part 1)

Video transcript

Let's attempt to take
the antiderivative, or the indefinite integral,
of x to the third times the square root of 9
minus x squared dx. And you might attempt
to do something like use substitution,
but you'll find that you're not
getting very far. And we have a big clue here. We have something
of the form-- so 9 minus x squared, that can
be viewed as the same thing as 3 squared minus x squared. And any time you have the form
a squared minus x squared, it might be useful to
make the substitution that x is equal to a sine theta. Now, why would that be? Well, then a squared
minus x squared would become a squared
minus a squared sine squared theta, which is the same
thing as a squared times 1 minus sine squared theta. And I think you see
where this is going. This is a squared
times cosine squared of theta, which might be
a useful simplification. So let's do it over here. In this case, our a is 3. So let's make the
substitution that x is going to be equal to a sine
theta, or three times sine of theta. And then we're
going to also have to figure out what
dx is equal to. So if you take the
derivative, we will get dx. We could have dx d theta
is equal to 3 cosine theta. Or if we wanted to write
it in differential form, we could write that
dx is equal to 3 times cosine theta d theta. This is just the derivative
of this with respect to theta. And we're ready to
substitute back. Our original expression
now becomes-- I'll write it in that
original green-- 3 sine theta to the third power, which
is the same thing as 27 sine-- actually,
let me color code it just so you know
what parts I'm doing. So this part right over
here, x to the third, is now going to become 27 sine
to the third power of theta, or sine theta to
the third power. And then all of
this business, this is going to be the square
root of 9 minus x squared, so minus 9 sine squared theta. And then dx-- let me do
this in a new color-- dx right over here is going
to be equal to-- that's not a new color. dx is going to be equal
to all of this business, so times 3 cosine theta d theta. And now let's see
if we can simplify this business a little bit. Let me do this over to the side. This thing right over here
can be written as 9 times 1 minus sine squared
theta, which is equal to the square root of
9 times cosine squared theta. And we can assume that
cosine theta is positive, as we did in the last video. And so this is going to be equal
to 3 cosine theta in orange. So this right over
here is 3 cosine theta. And so what does
this simplify to? We have a 27 times 3 is 81
times 3 is going to be 243. So this is going to
be 243-- I'll put it out front-- times the
integral of-- let's see, we're going to have
sine cubed theta. And then you're going to have
cosine theta times cosine theta, or we could say
cosine squared theta. That's this term right over
here and this term right over there, and of
course, d theta. I think I've taken
care of everything. So it might not look
like I've simplified it a lot, because, hey,
look, this still doesn't seem like a trivially
easy problem to solve, but we are getting closer. Now this turns into just a
classic u substitution problem. And it's not obvious just yet. There's actually a little
bit of a layer of a technique to figure out first. How do you do u substitution
right over here? And the key when you have
powers of trig functions, especially when you
have one of them as an odd power,
what you want to do is separate one of
those odd powers out so you can kind of construct
a u substitution problem. So let's do that. So this is going to
be equal to 243 times the integral of--
sine cubed theta I can rewrite as sine
of theta-- actually, let me write it
this way-- as sine squared theta cosine
squared theta. And then I still have 1 sine
theta right over here, d theta. And what I'm trying to do
is turn this expression into something where I
can do u substitution, and as you could
imagine, maybe where the du has to deal with
sine theta d theta. And it would if I can get my
u being equal to cosine theta. Then my du is going to be
negative sine theta d theta. So let me see if I can do that. So if I say that
sine squared theta is the same thing as 1
minus cosine squared theta, then this whole thing becomes
243 times the integral of 1 minus cosine squared theta
times cosine squared theta times sine theta d theta. And I want to be very
clear what I did over here. So we used some
trig substitution to get to this point
right over here. And at this point, I took
one of the sine thetas out-- I separated
it right over here-- and then I converted
this to an expression in terms of cosine theta. Now, the whole reason why I
did this is right over here I have a function of
cosine theta, and then I have something that's pretty
close to the derivative of cosine theta right over here. So this is now ripe
for u substitution. So let's do u substitution. If I have a function
of something and then I have this
derivative, maybe u should be equal
to that something. So let me set u as being
equal to cosine of theta. Then du is going to be equal to
negative sine of theta d theta. Well, I have a
sine theta d theta. I can multiply that
times a negative as long as I put a
negative out here. I'm multiplying by
a negative twice. I'm not changing the value. So notice, now this
right over here is du, and this right over
here is a function of u. So let's write it that way. All of this business is going to
be equal to negative 243 times the integral of 1 minus u
squared times u squared. And then this right over
here is just our du. So this is just du. Now, this is pretty
straightforward. We can just multiply
our u out, and this will become negative 243. So this is all equal
to negative 243 times the indefinite
integral of u squared minus u to the fourth--
I'm just distributing the u squared-- du. Now, this is pretty
straightforward to take the antiderivative of. This is negative 243 times the
antiderivative of u squared is u to the third over 3. Antiderivative of
u to the fourth is u to the fifth over 5. And of course, we're going
to have a plus c out here. And just so that we can
get rid of this negative, we can swap this. We can distribute
the negative sign. So this one becomes negative. That one becomes positive. And we get all of this business
as being equal to 243 times u to the fifth over 5 minus u
to the third over 3 plus c. And you might say, wow,
finally, we are done. But we aren't done. We have everything in terms of
u while our original integral was in terms of x. So the next video, we're going
to unwind all the substitution. We're going to try to take
this expression right over here and write it in terms of x. So we're going to have to
go from u to theta to x, because we've done two
rounds of substitutions.