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$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x=$
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1948 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{20}$
(C)
$\begin{array}{l}
\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x...(1)
\\
=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)+\cot \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)}{\tan \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)}\right] d x=\int_{\frac{\pi}{5}}^{10} \frac{3 \pi}{\tan \left(\frac{\pi}{2}-x\right)+\cot \left(\frac{\pi}{2}-x\right)} \tan \left(\frac{\pi}{2}-x\right) \\
=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x+\cot x} d x...(2)
\end{array}$
Equation (1) + (2) gives
$\begin{aligned} 2 I &=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} d x=[x]_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}=\left(\frac{3 \pi}{10}-\frac{\pi}{5}\right)=\frac{\pi}{10} \\ \therefore I &=\frac{\pi}{20} \end{aligned}$
$\begin{array}{l}
\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x...(1)
\\
=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)+\cot \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)}{\tan \left(\frac{3 \pi}{10}+\frac{\pi}{5}-x\right)}\right] d x=\int_{\frac{\pi}{5}}^{10} \frac{3 \pi}{\tan \left(\frac{\pi}{2}-x\right)+\cot \left(\frac{\pi}{2}-x\right)} \tan \left(\frac{\pi}{2}-x\right) \\
=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x+\cot x} d x...(2)
\end{array}$
Equation (1) + (2) gives
$\begin{aligned} 2 I &=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} d x=[x]_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}=\left(\frac{3 \pi}{10}-\frac{\pi}{5}\right)=\frac{\pi}{10} \\ \therefore I &=\frac{\pi}{20} \end{aligned}$
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