Question

${\int }_0^{\pi /2}\log \left(\sin x\right)dx=$

• A. $-\left(\frac{\pi }{2}\right)\log 2$
• B. $\pi \log \frac{1}{2}$
• C. $-\pi \log \frac{1}{2}$
• D. $\frac{1}{2}\log 2$

Answer 1

The correct option is A $-\left(\frac{\pi }{2}\right)\log 2$

$I={\int }_0^{\pi /2}\log \left(\sin x\right)dx={\int }_0^{\pi /2}\log \left(\cos x\right)dx\Rightarrow 2I={\int }_0^{\pi /2}\log \left(\sin x\cos x\right)dx={\int }_0^{\pi /2}\log \left(\sin 2x\right)dx-{\int }_0^{\pi /2}\log \left(2\right)dx$
$=\frac{1}{2}{\int }_0^{\pi }\log \left(\sin t\right)dt-\frac{\pi }{2}\log 2[\text{Putting in first integral}2x=t]=\frac{1}{2}\cdot 2{\int }_0^{\pi /2}\log \sin tdt-\frac{\pi }{2}\log 2[\because {\int }_0^{2a}f\left(x\right)dx=2\cdot {\int }_0^{a}f\left(x\right)dx\text{if}f(2a-x)=f\left(x\right)]\Rightarrow 2I-I=-\frac{\pi }{2}\log 2\Rightarrow I=-\frac{\pi }{2}\log 2.[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(t\right)dt]$