Question

Solve: $\int {\tan }^{-1}\sqrt{\frac{1-x}{1+x}}dx$

Answer 1

Put $x=cos2t$,
=> $dx=-2sin2tdt$
=> $\int -2ta{n}^{-1}\sqrt{\frac{1-cos2t}{1+cos2t}}sin2tdt$
=> $\int -2ta{n}^{-1}\sqrt{\frac{2si{n}^{2}t}{2co{s}^{2}t}}sin2tdt$
=> $\int -2ta{n}^{-1}\left(tant\right).sin2tdt$
=> $-2\int t.sin2tdt$
Solving this using By Parts, we get
=> $-2[\frac{t^{2}sin2t}{2}+\int \frac{cos2t}{2}]dt$
=> $-[t^{2}sin2t+\frac{sin2t}{2}]+C$
Put $t=co{s}^{-1}\frac{x}{2}$,
=> $-[co{s}^{-1}{\frac{x}{2}}^2.sin(co{s}^{-1}x)+\frac{sin\left(co{s}^{-1}x\right)}{2}]+C$