Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Evaluate $\int_{0}^{\frac{\pi}{2}} \log \sin x d x$
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Answer
Let $I=\int_{0}^{\frac{\pi}{2}} \log \sin x d x$ $I=\int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\frac{\pi}{2}} \log \cos x d x$ Adding the two values of I, we get $2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x$ = $\int_{0}^{\frac{\pi}{2}}(\log \sin x \cos x+\log 2-\log 2) d x$ (by adding and subtracting log 2) = $\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x$ Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when $x=\frac{\pi}{2}$, t = $\pi$ Therefore, 2I = $\frac{1}{2} \int_{0}^{\pi} \log \sin t d t-\frac{\pi}{2} \log 2$ = $\frac{2}{2} \int_{0}^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{2} \log 2$ = $\int_{0}^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{2} \log 2$ ((by changing variable t to x) = $I-\frac{\pi}{2} \log 2$ Hence, $\int_{0}^{\frac{\pi}{2}} \log \sin x d x=\frac{-\pi}{2} \log 2$
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