Question

Question #7c0a6

Answer 1

`pi^2/8`

Explanation:

This integral can be approached by considering a suitable u substitution

Let `u=sin^-1x`
We can differentiate this implicitly;
`sinu = x`
`cosu*(du)/dx = 1`
`(du)/dx = 1/cosu`

Now considering how `sinu = x` and how `cos^2x+sin^2x=1`
So hence `cosu = root2 (1-x^2)`

Hence `du = (dx)/(root2 ( 1-x^2))`

Hence the intergal becomes `int u du`
`Hence; (1/2)u^2 + c`

Hence the antiderivative is `1/2 (sin^-1x)^2 +c`

Hence evaluating from 0 to 1 we get;

`1/2 (pi/2)^2 = pi^2/8`