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

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Answer 1 · 2015-08-17T09:58:28

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

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

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

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