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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

 (17.43)

so

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

17.44   Example. Find .

Let . Then , and

 (17.45)

Thus

To find , I will integrate by parts. Let

Hence,

Hence

i.e.,
 (17.46)

Hence

By (17.45) we have and . Thus

17.47   Example. In the process of working out the last example we found using integration by parts. Here is a different tricky way of finding the same integral [32].

17.48   Example. Find .

Let . Then and

Thus

17.49   Exercise. Find the following antiderivatives:
a)

b)

c)

d)
A

17.50   Example (Area of a circular sector) Let be a positive number, and let be a number in . Let , and let . Let be the circular sector bounded by the positive -axis, the segment , and the circle .
The equation for is

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 .

Next: 17.6 Substitution in Integrals Up: 17. Antidifferentiation Techniques Previous: 17.4 Integration by Substitution   Index
Ray Mayer 2007-09-07