17.5 Trigonometric Substitution

Integrals of the form
and
often arise in applications. There is a special trick for dealing with such
integrals. Since

we can write

If we now make the substitution

then we find , or

Now

so

(Since we have and the square root is positive.) Thus

Often this last antiderivative can be found. If

then by the ritual (17.37)

The ritual to apply when using this method for finding is:

Let . Then
, and

so

There is a similar ritual for integrals of the form (Here we will just describe the ritual).

Let . Then
and

Observe that in equation (17.43) we are assuming that , so , so , and the sign of the square root is correct.

Let
. Then
, and

To find , I will integrate by parts. Let

Hence,

Hence

i.e.,

Hence

By (17.45) we have and . Thus

Let
. Then
and

Thus

and the equation for the upper semicircle is

Hence

where

i.e.

In exercise 17.49A.a you showed that

so

By using symmetry arguments, you can show that this formula actually holds for .