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Question: Answered & Verified by Expert
$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=$
MathematicsDefinite IntegrationMHT CETMHT CET 2020 (16 Oct Shift 1)
Options:
  • A 1
  • B $\frac{\pi}{4}$
  • C 0
  • D $\frac{\pi}{8}$
Solution:
1408 Upvotes Verified Answer
The correct answer is: 0
(C)
$\int_{0}^{\frac{\pi}{2}} \log \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}} d x$
$=\int_{0}^{\pi / 2} \log \left(\frac{\sin x}{\cos x}\right) d x \Rightarrow \int_{0}^{\pi / 2} \log (\tan x) d x$...(1)
$=\int_{0}^{\pi / 2}\left[\tan \left(\frac{\pi}{2}-x\right)\right] d x=\int_{0}^{\pi / 2} \log \cot x d x$...(2)
Eq. (1) $+(2)$ gives
$\begin{aligned}
2 \mathrm{I} &=\int_{0}^{\pi / 2}[\log (\tan \mathrm{x})+\log (\cot \mathrm{x})] \mathrm{dx}=\int_{0}^{\pi / 2} \log [\tan \mathrm{x} \cot \mathrm{x}] \mathrm{d} \mathrm{x} \\
&=\int_{0}^{\pi / 2} \log 1 \mathrm{dx}=0 \Rightarrow \mathrm{I}=0
\end{aligned}$

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