Welcome to an example
of integration using trig substitution.
Before trying trig substitution though,
we should make sure that basic u-substitution won't work.
So looking at our integral, notice how we might try
to let u equal the integrand of 36 minus x squared.
But notice if we do this,
differential u would be equal to negative two x dx,
which does not fit the form of the integral.
But because the integral does contain
the square root of 36 minus x squared,
which fits the form of the square root
of a squared minus x squared,
where a squared is a constant,
we can use trig substitution and let x equal a sine theta.
So we need to be able to recognize the form
where the square root indicates
which trig substitution we will use.
For example, if the integral contained
the square root of a squared plus x squared,
we would let x equal a tangent theta.
And if the integral involved the square root
of x squared minus a squared,
we'd let x equal a secant theta.
So because basic u-substitution failed
and it fits this form here,
where a squared is equal to 36 and therefore a equals six,
we will let x equal six sine theta.
And therefore differential x or dx
is equal to six cosine theta d theta.
Before we go on, let's sketch
a reference triangle for angle theta.
Notice if x equals six sine theta,
then sine theta equals x divided by six.
So if we sketch a right triangle
and call this angle theta,
we can label the opposite side x and the hypotenuse six,
and therefore this other leg would be equal
to the square root of six squared minus x squared,
or the square root of 36 minus x squared
using the Pythagorean theorem.
Now let's go ahead and perform our substitution.
We would have the square root
of 36 minus x squared.
Well if x is six sine theta,
x squared is 36 sine squared theta.
dx is equal to six cosine theta d theta.
And the denominator is four x squared,
which would be four times 36 sine squared theta.
Now let's begin to simplify.
Let's first take a look at this square root.
If we factor underneath the square root,
we can factor out 36,
which would give us the square root of 36
times the quantity one minus sine squared theta.
And the square root of 36 is equal to six.
So it'd have the integral of six.
And then one minus sine squared theta
is equal to cosine squared theta.
And the square root of cosine squared theta is cosine theta.
So we have six cosine theta
times six cosine theta, d theta,
divided by four times 36 times sine squared theta.
And since six times six is 36,
these two sixes simplify to one, and so does this 36.
So we can factor out 1/4,
since the four is in the denominator.
And we'd have the integral
of cosine squared over sine squared.
Notice in this form, we might think to try u-substitution,
where if we let u equal sine theta.
But notice that differential u
would be cosine theta d theta,
which does not fit the form of the integral
because we have cosine squared theta.
So let's go ahead and write cosine squared theta
divided by sine squared theta as cotangent squared theta.
Now in this form, we still don't have
a basic integration formula that
we can use for cotangent squared theta.
So now we'll perform our substitution
using a Pythagorean identity.
So to review, looking at this identity here,
we can replace cotangent squared theta
with cosecant squared theta minus one.
And the reason this helps is there's an integration formula
for cosecant squared theta, as we see here.
So we can write this as 1/4 times the integral
of cosecant squared theta minus one, d theta.
And now we can integrate.
So we'll have 1/4 times the quantity.
Well the integral of cosecant squared theta
is equal to negative cotangent theta
as we saw on the previous slide.
The integral of one would be theta.
Then we have plus C.
Let's go ahead and factor out negative one here.
That would give us negative 1/4
times cotangent theta
plus theta plus C.
So here's our antiderivative in terms of theta,
but we need this in terms of x.
So now we'll perform a substitution
for cotangent theta as well as theta,
which will require an inverse trig function.
So going back to our reference triangle,
cotangent theta would be equal to the ratio
of the adjacent side to the opposite side,
which should be the square root
of 36 minus x squared divided by x.
Let's perform that substitution first.
And then for theta, we'll use an inverse trig function.
So looking at this equation here,
if sine theta equals x divided by six,
then theta must be equal to inverse sine
or arcsine of x divided by six.
So while we could use a different inverse trig function,
let's use arcsine.
So we'd have plus arcsine,
x divided by six plus C.
And I think we'll go ahead and leave it in this form here.
This would be our anti-derivative in terms of x.
I hope you found this explanation helpful.