# Integrate sin(tan^-1x)dx please simplify and then solve

Anshuman Guru
40 Points
5 years ago
$\int sin(tan^{-1}x)dx$
Let $tan^{-1}x=\theta$
=> tan θ = x
=> $\sec \theta =\sqrt{1+x^2}$
=> $\cos \theta =1/\sqrt{1+x^2}$
=> $\sin \theta =x/\sqrt{1+x^2}$
=> $\theta =\sin^{-1}(x/\sqrt{1+x^2})$
I= $\int \sin(\sin^{-1}(x/\sqrt{1+x^2}))dx$
= $\int (x/\sqrt{1+x^2}))dx$
let x2=t
=> 2xdx=dt
=> xdx=dt/2
I= $\frac{1}{2}\int \frac{dt}{\sqrt{1+t}}$
= $\sqrt{1+t} + c$
= $\sqrt{1+x^{2}} + c$
just convert tan-1 x to sin-1 x using angles or you can just do it by triangle method..... then as the derivative of denominator is present in numerator so either take x^2 or 1+x^2 as t and substitute accoringly
Hope this helps :-)