By using the properties of definite integrals | vedclass

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(A) Let $I = \int_{0}^{\frac{\pi}{2}}(2 \log \sin x - \log \sin 2x) dx$.Using the property $\sin 2x = 2 \sin x \cos x$,we get:$I = \int_{0}^{\frac{\pi

Vedclass · Accepted Answer

$-\frac{\pi}{2} \log 2$

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(A) Let $I = \int_{0}^{\frac{\pi}{2}}(2 \log \sin x - \log \sin 2x) dx$.Using the property $\sin 2x = 2 \sin x \cos x$,we get:$I = \int_{0}^{\frac{\pi}{2}} \{2 \log \sin x - \log (2 \sin x \cos x)\} dx$Using $\log(abc) = \log a + \log b + \log c$:$I = \int_{0}^{\frac{\pi}{2}} \{2 \log \sin x - (\log 2 + \log \sin x + \log \cos x)\} dx$$I = \int_{0}^{\frac{\pi}{2}} (\log \sin x - \log \cos x - \log 2) dx$ ...... $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:$I = \int_{0}^{\frac{\pi}{2}} (\log \cos x - \log \sin x - \log 2) dx$ ...... $(2)$Adding $(1)$ and $(2)$:$2I = \int_{0}^{\frac{\pi}{2}} (\log \sin x - \log \cos x - \log 2 + \log \cos x - \log \sin x - \log 2) dx$$2I = \int_{0}^{\frac{\pi}{2}} (-2 \log 2) dx$$2I = -2 \log 2 [x]_{0}^{\frac{\pi}{2}}$$2I = -2 \log 2 (\frac{\pi}{2})$$2I = -\pi \log 2$$I = -\frac{\pi}{2} \log 2$.

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$.

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