MHT CET · Maths · Definite Integration
\(\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=\)
- A 1
- B \(\frac{\pi}{4}\)
- C 0
- D \(\frac{\pi}{8}\)
Answer & Solution
Correct Answer
(C) 0
Step-by-step Solution
Detailed explanation
(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
\(2 \mathrm{I} =\int_{0}^{\pi / 2}[\log (\tan \mathrm{x})+\log (\cot \mathrm{x})] \mathrm{dx}=\int_{0}^{\pi / 2}\) \(\log [\tan \mathrm{x} \cot \mathrm{x}] \mathrm{d} \mathrm{x} \)
\( =\int_{0}^{\pi / 2} \log 1 \mathrm{dx}=0 \Rightarrow \mathrm{I}=0 \)
\(\int_{0}^{\frac{\pi}{2}} \log \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}} d x\)
\(=\int_{0}^{\pi / 2} \log \left(\frac{\sin x}{\cos x}\right) d x \Rightarrow \int_{0}^{\pi / 2} \log (\tan x) d x\)...(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
\(2 \mathrm{I} =\int_{0}^{\pi / 2}[\log (\tan \mathrm{x})+\log (\cot \mathrm{x})] \mathrm{dx}=\int_{0}^{\pi / 2}\) \(\log [\tan \mathrm{x} \cot \mathrm{x}] \mathrm{d} \mathrm{x} \)
\( =\int_{0}^{\pi / 2} \log 1 \mathrm{dx}=0 \Rightarrow \mathrm{I}=0 \)
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