Question

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

Answer 1

`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`