Using integral _ 0 ^ / 2 ( x) d x =- _ 0 ^ / 2 ( x) d x=- 2 2 _ 0 ^ / 2 ( x) d x=0 and _…
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
Using integral \(\int_{0}^{\pi / 2} \ln (\sin x) d x\)
\(=-\int_{0}^{\pi / 2} \ln (\sec x) d x=-\frac{\pi}{2} \ln 2\)
\(\int_{0}^{\pi / 2} \ln (\tan x) d x=0\) and \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x=\frac{\pi}{8} \ln 2\)
Evaluate \(\int_{-\pi / 4}^{\pi / 4} \ln (\sin x+\cos x) d x=\)
- \(\frac{\pi \ln 2}{2}\)
- \(\frac{-\pi \ln 2}{4}\)
- \(\pi \ln 2\)
- \(0\)
Answer
(B) - 2 4
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