Question

Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.

Answer 1

Let I = `int (sin^-1x)/(1 - x^2)^(3//2) dx`

Consider t = sin^–1 x

`dt/dx = 1/sqrt(1 - x^2)`

∴ I = `int (t.dt)/((1 - x^2))`

= `int (t.dt)/((1 - sin^2t))`

= `int (t.dt)/(cos^2t)`

= `int t . sec^2 t  dt`

On integrating by parts

= `t int sec^2t.dt - int {(d(t))/dt int sec^2 t}dt`

= `t tan t - int 1.tan t  dt`

= t tan t – log sec t + C

= sin^–1x tan [sin^–1x] – log sec [sin^–1x] + C