int_0^{pi /2} log(sin x) , dx = | vedclass

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(A) Let $I = \int_0^{\pi /2} \log(\sin x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have $I = \int_0^{\pi /2} \log(\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int_0^{\pi /2} \log(\sin x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have $I = \int_0^{\pi /2} \log(\cos x) \, dx$. Adding these two equations: $2I = \int_0^{\pi /2} (\log(\sin x) + \log(\cos x)) \, dx = \int_0^{\pi /2} \log(\sin x \cos x) \, dx$. Multiplying and dividing by $2$ inside the log: $2I = \int_0^{\pi /2} \log(\frac{\sin 2x}{2}) \, dx = \int_0^{\pi /2} \log(\sin 2x) \, dx - \int_0^{\pi /2} \log 2 \, dx$. $2I = \int_0^{\pi /2} \log(\sin 2x) \, dx - \frac{\pi}{2} \log 2$. Let $2x = t$,then $2 \, dx = dt$. When $x=0, t=0$ and when $x=\pi/2, t=\pi$. $2I = \frac{1}{2} \int_0^{\pi} \log(\sin t) \, dt - \frac{\pi}{2} \log 2$. Using $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: $2I = \frac{1}{2} \cdot 2 \int_0^{\pi /2} \log(\sin t) \, dt - \frac{\pi}{2} \log 2$. $2I = I - \frac{\pi}{2} \log 2$. $I = -\frac{\pi}{2} \log 2$.

$\int_0^{\pi /2} \log(\sin x) \, dx = $

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