JEE Integration

ILATE RULE

$x\log(\sin x)\bigg|_0^{\pi/2} - \int\limits_0^{\pi/2}\log(\sin x)\mathrm dx= -\left(-\frac{\pi}{2}\log2\right) = 1.08879$

Using integration by parts with $u = x$ and $dv = \cot x dx$ , we have: $I = x \ln(\sin x)|_0^{\pi/2} - \int_0^{\pi/2} \ln(\sin x) dx$

Let $I_1 = \int_0^{\pi/2} \ln(\sin x) dx$ , and note that:

$I_1 = \int_0^{\pi/2} \ln(2 \sin \left( \dfrac{x}{2} \right) \cos \left(\dfrac{x}{2} \right) dx \\ = \int_0^{\pi/2} \ln(2) dx + \int_0^{\pi/2} \ln \left( \sin \left( \dfrac{x}{2} \right) \right) dx + \int_0^{\pi/2} \ln \left( \cos \left( \dfrac{x}{2} \right) \right) dx \\ = \dfrac{\pi ln(2)}{2} + I_2 + I_3$

Let $x = 2u$ so that $I_2 + I_3 = 2 \int_0^{\pi/4} \ln (\sin u) du + 2 \int_0^{\pi/4} \ln(\cos u) du$

Let $u = \dfrac{\pi}{2} - t$ so that $2 \int_0^{\pi/4} \ln(\cos u) du = 2 \int_{\pi/4}^{\pi/2} \ln(\sin t) dt$ , and so $I_1 = \dfrac{\pi ln(2)}{2} + 2 \int_{0}^{\pi/2} \ln(\sin x) dx = \dfrac{\pi ln(2)}{2} + 2I_1$ $I_1 = -\dfrac{\pi ln(2)}{2}$

Finally, $I = \displaystyle \lim_{b \to 0} b \ln(\sin b) - \left( -\dfrac{\pi ln(2)}{2} \right) = 0 + \dfrac{\pi ln(2)}{2} \sim 1.088$ , where the limit is evaluated using L’Hospital’s Rule.