Question

\[\int\limits_0^\pi x \log \sin x\ dx\]

Answer 1

\[\int_0^\pi x \log \sin x\ d x\]
\[Let I = \int_0^\pi x \log\left( \sin x \right) d\ x . . . . . (i)\]
\[ I = \int_0^\pi \left( \pi - x \right) \log \sin\left( \pi - x \right) d x\]
\[ I = \int_0^\pi \left( \pi - x \right) \log\left( \sin x \right) dx . . . . . (ii)\]
\[\text{Adding (i) and (ii)}\]
\[2I = \pi \int_0^\pi \log \sin x\ d x\]
\[ = 2\pi \int_0^\frac{\pi}{2} \log \sin x\ d x\]
\[ I = \pi \int_0^\frac{\pi}{2} \log \sin x\ d x . . . . . (iii)\]
\[Let\ \int_0^\frac{\pi}{2} \log \sin x dx = I_2 \]
\[ I_2 = \int_0^\frac{\pi}{2} \log \sin\left( \frac{\pi}{2} - x \right) dx\]
\[ = \int_0^\frac{\pi}{2} \log \cos x dx\]
\[2 I_2 = \int_0^\frac{\pi}{2} \left( \log \sin x + \log \cos x \right) dx\]
\[ = \int_0^\frac{\pi}{2} \log\left( \sin x \cos x \right) dx\]
\[ = \int_0^\frac{\pi}{2} \log\left( \sin2x \right) dx - \int_0^\frac{\pi}{2} \log 2 dx\]
\[Let\ 2x = t\]
\[2dx = dt\]
\[when, \]
\[x = 0 \Rightarrow t = 0\]
\[x = 0 \Rightarrow t = \pi\]
\[2 I_2 = \frac{1}{2} \int_0^\pi \log \left( \sin t \right) dt - \frac{\pi}{2}\log 2\]
\[2 I_2 = \frac{2}{2} \int_0^\frac{\pi}{2} \log \left( \sin t \right) dt - \frac{\pi}{2}\log 2\]
\[2 I_2 = I_2 - \frac{\pi}{2}\log 2\]
\[ I_2 = - \frac{\pi}{2}\log 2\]
\[From \left( iii \right), \]
\[ I = \pi \int_0^\frac{\pi}{2} \log\ sinx\ dx = \pi I_2 \]
\[I = \pi\left( - \frac{\pi}{2}\log 2 \right)\]
\[I = \frac{- \pi^2 \log 2}{2}\]