Question

Prove that ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$ and hence evaluate
${\int }_0^{\pi /2}(2\log \sin x-\log \sin 2x)$

Answer 1

Put $x=a-t$
$dx=-dt$
when $x=0;t=a$
$x=a;t=0$
LHS ${\int }_0^{a}f\left(x\right)dx={\int }_a^{0}f(a-t)(-dt)={\int }_0^{a}f(a-t)dt={\int }_0^{a}f(a-x)dx$
$I=2{\int }_0^{\pi /2}\log \left(\sin x\right)dx-{\int }_0^{\pi /2}\log \left(2\sin x\cos x\right)dx$
$=2{\int }_0^{\pi /2}\log \left(\sin x\right)dx-{\int }_0^{\pi /2}\log 2dx-{\int }_0^{\pi /2}\log \left(\sin x\right)dx-{\int }_0^{\pi /2}\log \left(\cos x\right)dx$
$={\int }_0^{\pi /2}\log \left(\sin x\right)dx-{\int }_0^{\pi /2}\log \left(\cos (\frac{\pi }{2}-x)\right)dx-\left(\log 2\right)\frac{\pi }{2}dx$
$\Rightarrow I=-\frac{\pi }{2}\log 2=\frac{\pi }{2}\log \frac{1}{2}$