Evaluate: `int(sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))\ dx`

Explanation

We have
`I=int({sin^(-1)sqrt(x)-((pi)/(2)-sin^(-1)sqrt(x))})/(((pi)/(2)))dx[becausesin^(-1)sqrt(x)+cos^(-1)sqrt(x)=(pi)/(2)]`
`=(2)/(pi)intsin^(-1)sqrt(x)dx-intdx=(4)/(pi)intsin^(-1)sqrt(x)dx-x+C`. . . . (i)
Putting `x=sin^(2)thetaanddx=2sinthetacosthetad theta=sin2theta`, we get
`intsin^(-1)sqrt(x)dx=intunderset(I)(theta)underset(" "II)(sin)2thetad theta`
`=theta((-cos2theta)/(2))-int1*((-cos2theta))/(2)d theta" [integrating by parts]"`
`=-(theta)/(2)cos2theta+int(1)/(2)cos2thetad theta`
`=-(1)/(2)thetacos2theta+(1)/(4)sin2theta`
`=-(1)/(2)theta(1-2sin^(2)theta)+(1)/(2)sinthetasqrt(1-sin^(2)theta)`
`=-(1)/(2)sin^(-1)sqrt(x)(1-2x)+(1)/(2)sqrt(x)*sqrt(1-x)`. . . .(ii)
From (i) and (ii), we get
`I=(4)/(pi)*{-(1)/(2)sin^(-1)sqrt(x)(1-2x)+(1)/(2)sqrt(x)sqrt(1-x)}-x+C`
`:.I=(-2)/(pi)sin^(-1)sqrt(x)(1-2x)+(1)/(pi)sqrt(x)sqrt(1-x)-x+C`.