$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{ an x}{ an x+\cot x} ight] d x=$

Question

$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{	an x}{	an x+\cot x}ight] d x=$

Aicharya · Accepted Answer

$\begin{array}{l} \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{	an x}{	an x+\cot x}ight] d x...(1) \ =\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{	an \left(\frac{3 \pi}{10}+\frac{\pi}{5}-xight)+\cot \left(\frac{3 \pi}{10}+\frac{\pi}{5}-xight)}{	an \left(\frac{3 \pi}{10}+\frac{\pi}{5}-xight)}ight] d x=\int_{\frac{\pi}{5}}^{10} \frac{3 \pi}{	an \left(\frac{\pi}{2}-xight)+\cot \left(\frac{\pi}{2}-xight)} 	an \left(\frac{\pi}{2}-xight) \ =\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{	an x+\cot x} d x...(2) \end{array}$ Equation (1) + (2) gives $\begin{aligned} 2 I &=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} d x=[x]_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}=\left(\frac{3 \pi}{10}-\frac{\pi}{5}ight)=\frac{\pi}{10} \ 	herefore I &=\frac{\pi}{20} \end{aligned}$

$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x=$

Solution