Question

${\int }_0^{\pi /2}\log \sin xdx=\frac{\pi }{k}\log \frac{1}{2}.$ Find the value of $k$.

Answer 1

Let $I={\int }_0^{\pi /2}\log \sin xdx$

By integration property
${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\therefore I={\int }_0^{\pi /2}\log \sin (\pi /2-x)dx={\int }_0^{\pi /2}\log \cos xdx$

$\therefore 2I={\int }_0^{\pi /2}(\log \sin x+\log \cos x)dx$

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

$={\int }_0^{\pi /2}\log \frac{\sin 2x}{2}dx.$

$={\int }_0^{\pi /2}\log \sin 2xdx-{\int }_0^{\pi /2}\log 2dx.$

Substitute $2x=t$ in the 1st. $\therefore 2dx=dt$ and limits become 0 to $\pi .$

$\therefore 2I=\frac{1}{2}{\int }_0^{\pi }\log \sin tdt-{\left[x\log 2\right]}_0^{\pi /2}$

Now apply Prop. in 1st. $\because f(2a-x)=f\left(x\right).$

$2I=\frac{1}{2}.2{\int }_0^{\pi /2}\log \sin tdt-\frac{\pi }{2}\log 2$

or $2I={\int }_0^{\pi /2}\log \sin xdx-\left(\pi /2\right)\log 2,$ by prop.

$2I=I+\frac{\pi }{2}\log \frac{1}{2}$

$\therefore I={\int }_0^{\pi /2}\log \sin xdx={\int }_0^{\pi /2}\log \cos xdx=\frac{\pi }{2}\log \frac{1}{2}$

Therefore $k=2$