Question

integration of tan^-1√1-cos2x/1+cos2x​

Answer 1

$\underline{\texttt{Solution :}}$

$\mathrm{Now,\:\int tan^{-1}\sqrt{\frac{1-cos2x}{1+cos2x}}\:dx}$

$\mathrm{=\int tan^{1}\sqrt{\frac{(sin^{2}x+cos^{2}x)-(cos^{2}x-sin^{2}x)}{(sin^{2}x+cos^{2}x)+(cos^{2}x-sin^{2}x)}}\:dx}$

$\mathrm{=\int tan^{-1} \sqrt{\frac{sin^{2}x+cos^{2}x-cos^{2}x+sin^{2}x}{sin^{2}x+cos^{2}x+cos^{2}x-sin^{2}x}}\:dx}$

$\mathrm{=\int tan^{-1}\sqrt{\frac{2sin^{2}x}{2cos^{2}x}}\:dx}$

$\mathrm{=\int tan^{-1}\sqrt{tan^{2}x}\:dx}$

$\mathrm{=\int tan^{-1}(tanx)\:dx}$

$\mathrm{=\int x\:dx}$

$\mathrm{=\frac{x^{2}}{2}+C}$

$\texttt{where C is integral constant}$

$\to \boxed{\small{\mathrm{\int tan^{-1}\sqrt{\frac{1-cos2x}{1+cos2x}}\:dx = \frac{x^{2}}{2}+C}}}$

$\texttt{which is the required integral}$

Answer 2

Answer is in attachment

thank you