How do you find the integral of #sqrt(9-x^2)dx#?

1 Answer
dani83
Aug 17, 2015

# int sqrt(9-x^2)dx = 9/2(sin^(-1) (x/3) + x/3sqrt(1-(x/3)^2)) + C #

Explanation:

# sin^2 A + cos^2 A = 1 #
# cos 2A = 2cos^2 A - 1 #
# sin 2A = 2sin Acos A #

Use substitution.
# x = 3 sin t #

# int sqrt(9-x^2)dx = int 3sqrt(1-sin^2 t) xx 3 cos t dt = 9 int cos^2 t dt #

# 9int cos^2 tdx = 9/2 int (1+cos 2t)dt = 9/2 (t + 1/2sin 2t) + C #

In terms of #x#:
# int sqrt(9-x^2)dx = 9/2(sin^(-1) (x/3) + x/3sqrt(1-(x/3)^2)) + C #

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