Find : `int(sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))dx ,x in [0,1]`

Explanation

`" Let " I= int (sin^(-1) sqrt(x) -cos^(-1) sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))dx`
we know that `sin^(-1) sqrt(x) +cos^(-1) sqrt(x) =(pi)/(2)`
`rArr cos^(-1) sqrt(x) =(pi)/(2) -sin^(-1) sqrt(x)`
`:. I= int (sin^(-1) sqrt(x) -((pi)/(2)-sin^(-1)sqrt(x)))/((pi)/(2))dx`
`=int (2sin^(-1)sqrt(x)-(pi)/(2))/((pi)/(2))dx`
`=(2)/(pi)int (2sin^(-1)sqrt(x)-(pi)/(2))dx`
`=(4)/(pi) intsin^(-1) sqrt(x) dx- int1l dx`
`=(4)/(pi) int sin^(-1) sqrt(x) dx- x`
`rArr I=(4)/(pi) I_(1) -x+C" '.....(1)`
`" where " I_(1)= int sin^(-1) sqrt(x) dx`
`" Let " sqrt(x) =t rArr x=t^(2) rArr dx=2tdt`
`I_(1)= int sin^(-1) t 2t dt = 2 int sin^(-1) t.t dt`
`=2 (sin^(-1) t.(t^(2))/(2)-int (1)/(sqrt(1-t^(2))).(t^(2))/(2)dt)`
`=t^(2) sin^(-1) t-int (-(1-t^(2))+1)/(sqrt(1-t^(2)))dt`
`=t^(2) sin^(-1) t+ int sqrt(1-t^(2))dt -int (1)/(sqrt(1-t^(2)))dt`
`=t^(2) sin^(-1) t+ (tsqrt(1-t^(2)))/(2)+(1)/(2) sin^(-1) t -sin^(-1)t`
`=(t^(2)-(1)/(2)) sin^(-1) t+(1)/(2) t^(sqrt(1-t^(2)))`
`=(1)/(2) [(2x-1) sin^(-1) sqrt(x) +sqrt(x) sqrt(1-x)]`
`=(1)/(2) [(2x-1) sin ^(-1) sqrt(x)+sqrt(x-x^(2))]`
Put the value of `I_(1)` equation (1)
` int (sin^(-1) sqrt(x) -cos^(-1) sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1) sqrt(x))dx`
`=(2)/(pi) [(2x-1)sin^(-1) sqrt(x)+sqrt(x-x^(2)]]-x+C`