Find the value of `int_0^1(sin^(-1)x)dx`

Answer

`(pi)/2 log_(e)2`

Explanation

`int_(0)^(1){(sin^(-1)x)//x}dx`
`[(sin^(-1))(logx)]_(0)^(1)-int_(0)^(1)1/(sqrt(1-x^(2)))logxdx`
`=0-lim_(xto0)(sin^(-1)x)/x(x log x)`
`-int_(0)^(pi//2)1/(sqrt(1-sin^(2)theta))(log sin theta) cos theta d theta`
`=-lim_(xto0) x log x -int_(0)^(pi//2) log sin theta d theta`
`=(pi)/2 log 2`