Question

How do I find the integral `intsin^-1(x)dx` ?

Answer 1

By integration by parts,
`int sin^{-1}xdx=xsin^{-1}x+sqrt{1-x^2}+C`

Let us look at some details.
Let `u=sin^{-1}x` and `dv=dx`.
`Rightarrow du={dx}/sqrt{1-x^2}` and `v=x`

By integration by parts,
`int sin^{-1}xdx=xsin^{-1}x-intx/sqrt{1-x^2}dx`

Let `u=1-x^2`. #Rightarrow {du}/{dx}=-2x Rightarrow dx={du}/{-2x}#

`intx/sqrt{1-x^2}dx=int x/sqrt{u}{du}/{-2x}=-1/2intu^{-1/2}du`
`=-u^{1/2}+C=-sqrt{1-x^2}+C`

Hence,
`int sin^{-1}xdx=xsin^{-1}x+sqrt{1-x^2}+C`