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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 . 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