Evaluate ∫limits_0^π x log sin x dx

The correct answer is Option (3) → $\frac{\pi^2}{2} \log \left( \frac{1}{2} \right)$

Let $I = \int\limits_{0}^{\pi} x \log \sin x \, dx \dots(i)$

$\Rightarrow I = \int\limits_{0}^{\pi} (\pi - x) \log \sin(\pi - x) \, dx$

$= \int\limits_{0}^{\pi} (\pi - x) \log \sin x \, dx \dots(ii)$

On adding Eqs. (i) and (ii), we get

$2I = \pi \int\limits_{0}^{\pi} \log \sin x \, dx \dots(iii)$

$2I = 2\pi \int\limits_{0}^{\pi/2} \log \sin x \, dx \quad \left[ ∵\int\limits_{0}^{2a} f(x) \, dx = 2 \int\limits_{0}^{a} f(x) \, dx \right]$

If $f(2a - x) = f(x)$

$\Rightarrow I = \pi \int\limits_{0}^{\pi/2} \log \sin x \, dx \dots(iv)$

Now, $I = \pi \int\limits_{0}^{\pi/2} \log \sin(\pi/2 - x) \, dx \dots(v)$

On adding Eqs. (iv) and (v), we get

$2I = \pi \int\limits_{0}^{\pi/2} (\log \sin x + \log \cos x) \, dx \quad [∵\sin(90 - x) = \cos x]$

$\Rightarrow 2I = \pi \int\limits_{0}^{\pi/2} \log \sin x \cos x \, dx$

$\Rightarrow 2I = \pi \int\limits_{0}^{\pi/2} \log \frac{2 \sin x \cos x}{2} \, dx$

$\Rightarrow 2I = \pi \int\limits_{0}^{\pi/2} (\log \sin 2x - \log 2) \, dx \quad [∵ \sin 2x = 2 \sin x \cdot \cos x]$

$\Rightarrow 2I = \pi \int\limits_{0}^{\pi/2} \log \sin 2x \, dx - \pi \int_{0}^{\pi/2} \log 2 \, dx$

Put $2x = t \Rightarrow dx = \frac{1}{2} dt$

As $x \to 0, \text{then } t \to 0$

and $x \to \frac{\pi}{2}, \text{then } t \to \pi$

$∴2I = \frac{\pi}{2} \int\limits_{0}^{\pi} \log \sin t \, dt - \frac{\pi^2}{2} \log 2$

$\Rightarrow 2I = \frac{\pi}{2} \int\limits_{0}^{\pi} \log \sin x \, dx - \frac{\pi^2}{2} \log 2$

$\Rightarrow 2I = I - \frac{\pi^2}{2} \log 2 \quad [\text{from Eq. (iii)}]$

$∴I = -\frac{\pi^2}{2} \log 2 = \frac{\pi^2}{2} \log \left( \frac{1}{2} \right)$