Question

${\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin xdx=$

• A. $-\mathrm{\pi }\log 2$
• B. $\mathrm{\pi }\log 2$
• C. $-\frac{\mathrm{\pi }}{2}\log 2$
• D. $\frac{\mathrm{\pi }}{2}\log 2$

Answer 1

The correct option is C

$-\frac{\mathrm{\pi }}{2}\log 2$

Explanation for Correct answer:

$I\text{=}{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin xdx\to \left(i\right)$

Also,

$$\begin{gathered} I\text{=}{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \left(\frac{\mathrm{\pi }}{2}-x\right)dx \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx\to \left(ii\right) \end{gathered}$$

Add $\left(i\right)$ and $\left(ii\right)$

$\begin{array}{rcl}2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos x+\log \sin xdx\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log (\cos x.\sin x)dx\left[\because \log a+\log b=\log ab\right]\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \frac{\sin 2x}{2}dx\left[\because \sin 2x=2\sin x\cos x\right]\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\log \sin 2x-\log 2\right)dx\left[\because \log a-\log b=\log \frac{a}{b}\right]\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin 2xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2dx\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin 2xdx-{\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}}\log 2\\ 2I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin 2xdx-\frac{\mathrm{\pi }}{2}\log 2\end{array}$

Let $2x=z\Rightarrow 2dx=dz$

Limits $x=0\Rightarrow z=0;x=\frac{\mathrm{\pi }}{2}\Rightarrow z=\mathrm{\pi }$

$\begin{array}{rcl}2I& =& \frac{1}{2}{\int }_0^{\mathrm{\pi }}\log \sin zdz-\frac{\mathrm{\pi }}{2}\log 2\\ 2I& =& \frac{1}{2}\times 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin zdz-\frac{\mathrm{\pi }}{2}\log 2\left[\because {\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx\text{if}f(2a-x)=f\left(x\right)\right]\\ 2I& =& I-\frac{\mathrm{\pi }}{2}\log 2\left[\text{from}\left(i\right)\right]\\ I& =& -\frac{\mathrm{\pi }}{2}\log 2\end{array}$

Hence, option (C) is the correct answer.