`intsin^(-1)(sqrt(x/(a+x)))dx` का मान ज्ञात कीजिए ।

Explanation

माना `x=a tan^2 theta`
`rArrdx=2a tan theta sec^2theta d theta`
`:. intsin^(-1)sqrt(x/(a+x))dx`
`=intsin^(-1)sqrt((atan^2theta)/(asec^2theta)).2a tan theta sec^2 theta d theta`
`=2aintsin^(-1)(sintheta)tanthetasec^2thetad theta`
`=2a intsin^(-1)(sin theta)tanthetasec^2thetad theta`
`=2a int thetatanthetasec^2thetad theta`
`=2a[theta.(tan^2theta)/2-int1.(tan^2theta)/2d theta] [:'int tan thetasec^2thetad theta=(tan^2theta)/2]`
`=a[thetatan^2theta-int(sec^2theta-1)d theta]`
`=a[thetatan^2theta-tantheta+theta]`
`=a[tan^(-1)sqrt(x/a).x/a-sqrt(x/a)+tan^(-1)sqrt(x/a)]+c (:' tan^2theta=x/a)`
`=(a+x)tan^(-1)sqrt(x/a)-sqrt(ax)+c`