Q. $\int_{0}^{1}{\log\left\{ \sin\left( \frac{\pi x}{2} \right) \right\}}\ dx$ is equal to
A) $- \frac{\pi}{2}\log 2$
B) $- \log 2$
C) $- \frac{2}{\pi}\log 2$
D) $\frac{\pi}{2}\log 2$
Explanation
(b)
Let $I = \int_{0}^{1}{\log\left\{ \sin\left( \frac{\pi x}{2} \right) \right\}}dx$
Put $\frac{\pi x}{2} = \theta\ \ \Rightarrow dx = \frac{2}{\pi}\ \ d\theta$
$\therefore\ I = \frac{2}{\pi}\ \int_{0}^{\frac{\pi}{2}}{\log{\sin\theta}}\ d\theta$
$= \frac{2}{\pi}\left( - \frac{\pi}{2}\log 2 \right) = - \log 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.