1.

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

Answer» Correct Answer - `(2)/(pi)[sqrt(x-x^(2))-(1-2x)"sin"^(-1)sqrt(x)]-x+c`
Let `I=int (sin^(-1)sqrt(x)-cos^(-1)sqrt(x))/(sin^(-1)sqrt(x)+cos^(-1)sqrt(x))dx`
`=int(sin^(-1)sqrt(x)-((pi)/(2)-sin^(-1)sqrt(x)))/((pi)/(2))dx`
`=(2)/(pi)int(2sin^(-1)sqrt(x)-(pi)/(2))dx=(4)/(pi)intsin^(-1)sqrt(x)dx - x + c" "...(i)`
Now, `int sin^(-1) sqrt(x)dx`
Put `x = sin^(2) theta rArr dx = sin 2 theta`
`=int theta*sin2 theta d theta =-(theta cos 2 theta)/(2)+int(1)/(2)cos 2 theta d theta`
`=-(theta)/(2)cos 2 theta + (1)/(4)sin 2 theta`
`=-(1)/(2)theta(1-2sin^(2)theta)+(1)/(2)sin theta sqrt(1-sin^(2)theta)`
`=-(1)/(2)sin^(-1)sqrt(x)(1-2x)+(1)/(2)sqrt(x)sqrt(1-x)" "...(ii)`
From Eqs. (i) and (ii),
`I=(4)/(pi)[-(1)/(2)(1-2x)sin^(-1)sqrt(x)+(1)/(2)sqrt(x-x^(2))]-x+c`
`=(2)/(pi)[sqrt(x-x^(2))-(1-2x)sin^(-1)sqrt(x)]-x+c`


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