`intsin^(- 1)sqrt(x/(a+x))dx`

Answer

N/a

Explanation

Let `I = intsin^(-1)sqrt((x)/(a+x))dx`
Put `x = atan^(2)x`
`rArr dx = 2 a tanthetasec^(2) thetad theta`
`rArr dx = 2atan theta sec^(2) theta d theta`
` :. I = int sin^(-1) sqrt((atan^(2) theta)/(a+a tan^(2) theta)) (2tan theta.sec^(2) theta) d theta`
`= 2aintsin^(-1)((tantheta)/(sectheta)) tan theta. sec^(2) theta d theta` ltbgt `= 2 a int sin^(-1)(sintheta) tan theta.sec^(2) theta d theta`
`=2aintunderset(I)(theta).tanthetaunderset(II)(sec^(2)theta)d theta`
`= 2a[theta.int tan theta. sec^(2) theta d theta- int (d/(d theta).inttantheta.sec^(2)thetad theta) d theta]`
`[{:(Put" "tantheta=t),(rArr sectheta.tantheta.d theta= dt),(rArr int tan thetasec^(2) theta d theta = intt dt):}]`
`= 2a[theta.(tan^(2)theta)/(2)-int(tan^(2)theta)/(2)d theta]`
` = atheta tan^(2) theta - a int (sec^(2) theta - 1) d theta`
`= a theta.tan^(2) theta- atan theta + a theta + C`
`= a [ (x)/(a) tan^(-1)sqrt((x)/(a)) + tan^(-1) sqrt((x)/(a))] + C`