First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{1} \log \sin (\frac{πx}{2}) dx$ is equal to

Moderate

A

log 2
B

-log 2
C

log 3
D

0

Solution

Let,
$\begin{aligned} I = \int_{0}^{1} \log \sin (\frac{πx}{2}) dx \\ put, \frac{πx}{2} = t \Rightarrow dx = \frac{2}{π} dt \end{aligned}$
$I = \frac{2}{π} \int_{0}^{π / 2} \log \sin tdt = \frac{2}{π} I_{1} - - - - - (i)$
where,
$\begin{matrix} I_{1} = \int_{0}^{π / 2} \log \sin tdt = \int_{0}^{π / 2} \log \sin (\frac{π}{2} - t) dt \\ = \int_{0}^{π / 2} \log \cos tdt \end{matrix}$
$\begin{array}{l} ∴ 2 I_{1} = \int_{0}^{π / 2} (\log \sin t + \log \cos t) dt \\ = \int_{0}^{π / 2} \log (\sin tcos t) dt = \int_{0}^{π / 2} \log (\frac{\sin 2 t}{2}) dt \\ = \int_{0}^{π / 2} (\log \sin 2 t - \log 2) dt \\ = \frac{1}{2} \int_{0}^{π} \log \sin zdz - \log 2 \int_{0}^{π / 2} dt \\ = \frac{1}{2} \cdot 2 \int_{0}^{π / 2} \log \sin zdz - \frac{π}{2} \log 2 \\ \Rightarrow 2 I_{1} = I_{1} - \frac{π}{2} \log 2 \Rightarrow I_{1} = - \frac{π}{2} \log 2 \end{array}$
From Eq. (i), $I = \frac{2}{π} (- \frac{π}{2} \log 2) = - \log 2$

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