For the indefinite integral below, choose which of the following substitutions would be most
helpful in evaluating the integral.
Enter the appropriate letter (A,B, or C) in each blank.
A. x= 3tan(theta)
B. x= 3sin(theta)
C. x= 3sec(theta)
integrate of (dx)/((9-x^2)^3/2)
Solution
Selecting the option B. [x = 3sin theta =>] [dx = 3 cos theta d theta] yields:
[int (3 cos theta d theta)/((9-9sin^2 theta)sqrt(9 - 9 sin theta))]
Factoring out 9 yields:
[int (3 cos theta d theta)/(27(1-sin^2 theta)sqrt(1 - sin^2 theta))]
Using the fundamental formula of trigonometry yields:
[1-sin^2 theta = cos^2 theta]
[int (3 cos theta d theta)/(27cos^3 theta) = (1/9) int(d theta)/(cos^2 theta)]
[(1/9) int (d theta)/(cos^2 theta) = (1/9) tan theta + c]
Substituting back [(sin^(-1)(x/3))] for [theta] yields:
[int (dx)/((9-x^2)^3/2)= (1/9) tan((sin^(-1)(x/3)))+ c]
Hence, evaluating the given integral yields [int (dx)/((9-x^2)^3/2)= (1/9) tan((sin^(-1)(x/3))) + c]
, using the option B. [x = 3sin theta] .