Question

${\int }_0^1\log \sin \left(\frac{\pi }{2}x\right)dx=k\log \frac{1}{2}.$ Find the value of $k$.

Answer 1

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