First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{π} x \log (\sin x) d x$ is (given that $\int_{0}^{π / 2} \log \sin x d x = - \frac{π}{2} \log 2)$

Easy

A

$- \frac{π^{2}}{2} \log 2$
B

$- \frac{π^{2}}{4} \log 2$
C

$- \frac{π^{2}}{8} \log 2$
D

none of these

Solution

$\begin{array}{l} I = \int_{0}^{π} x \log (\sin x) d x \\ = \int_{0}^{π} (π - x) \log [\sin (π - x)] d x \\ \Rightarrow 2 I = π \int_{0}^{π} \log \sin x d x \\ = 2 π [\int_{0}^{π / 2} \log \sin x d x] \\ \Rightarrow I = π \int_{0}^{π / 2} \log (\sin) d x = - \frac{π^{2}}{2} \log 2 \end{array}$

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