∫tan(sin^-1x)dx/
√ 1-x^2
let,tan(sin^-1x)=z
or, d/dx tan(sin^-1x)=dz/dx
or,sec^2(sin^-1x).1/√ 1-x^2=dz/dx
or, dx/√ 1-x^2=dz/1-tan^2(sin^-1x)
or,dx/√ 1-x^2=dz/1+z^2

∫tan(sin^-1x)dx/
√ 1-x^2=∫z.dz/1+z^2=1/1+z^2∫zdz=1/1+z^2.z^1+1/1+1+c=1/1+z^2.z^2/2+c=1/1+(tansin^-1x)2.(tansin^-1x)^2/2+c=tan(sin^-1x)^2/2(1+(tansin^-1x)^2+c
is this is right?

Question

∫tan(sin^-1x)dx/
√ 1-x^2
let,tan(sin^-1x)=z
or, d/dx tan(sin^-1x)=dz/dx
or,sec^2(sin^-1x).1/√ 1-x^2=dz/dx
or, dx/√ 1-x^2=dz/1-tan^2(sin^-1x)
or,dx/√ 1-x^2=dz/1+z^2

∫tan(sin^-1x)dx/
√ 1-x^2=∫z.dz/1+z^2=1/1+z^2∫zdz=1/1+z^2.z^1+1/1+1+c=1/1+z^2.z^2/2+c=1/1+(tansin^-1x)2.(tansin^-1x)^2/2+c=tan(sin^-1x)^2/2(1+(tansin^-1x)^2+c
is this is right?

austinl · Answer

Okay first off, wow. That is a mess.
You need to learn $\LaTeX$

It will make everything much easier to understand.

Take your first line for example.

$\Large{\int \frac{\tan{(\sin^{-1}{(x)})} dx}{\sqrt{1-x^2}}}$

anonymous · Answer

$u=\sin^{-1}(x)$ gets it

austinl · Answer

That is what I was thinking, a simple substitution.
Very nice.

anonymous · Answer

long as your recall 
$$\int \tan(u)du=\ln(\sec(u))$$ you are done in one step

anonymous · Answer

@radar

anonymous · Answer

@Rohitkhanna