ज्ञात करें `int (sin^(-1)x)/x^2dx`

Explanation

माना की `int (sin^(-1)x)/x^2dx`
`sin^(-1)x=theta` रखने पर `x=sin theta :. dx = cos theta d theta`
`:." "I=int ""theta/(sin^2 theta) cos theta d theta = int theta ( "cosec"theta * cot theta) d theta`
`= theta ( "-cosec"theta) - int 1("-cosec" theta) d theta` [ खंडनशाह विधि से स्मालकॉन करने पर ]
`=theta "cosec" theta + log | "cosec"theta- cot theta | +c`
`=-(sin^(-1)x)/x+log |1/x-sqrt(1-x^2)/x|+c=(sin^(-1)x)/x+log |(1-sqrt(1-x^2))/x|+c`