Question

$\int {\tan }^{-1}\left(\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right)dx$.

Answer 1

Solution $\int tan-1\left(\sqrt{\frac{1-cos2x}{1+cos2x}}\right)dx$

$cos2x=1-2si{n}^{2}x$

$2si{n}^{2}x=1-cos2x$

$cos2x=2co{s}^{2}x-1$

$2co{s}^{2}x=1+cos2x$

using this in the integral term

$\int ta{n}^{-1}\left(\sqrt{\frac{2si{n}^{2}x}{2co{s}^{2}x}}\right)dx$

$[\frac{sinx}{cosx}=tanx]$

$\int ta{n}^{-1}\sqrt{ta{n}^{2}x}dx$

$\int ta{n}^{-1}\left(tanx\right)dx$

$=\int dx=\frac{x^2}{2}Ans$

$[\int x^{n}dx=\frac{x^{n+1}}{n+1}]$