Question

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

Answer 1

Consider the given integral.

$I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx$ …….. (1)

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin (\frac{\pi }{2}-x)\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\ln \left(\cos x\right)dx$ ……. (2)

On adding equation (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)+\ln \left(\cos x\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\cos x\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\frac{2\sin x\cos x}{2}\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin 2x\right)dx-{\int }_0^{\frac{\pi }{2}}\ln \left(2\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin 2x\right)dx-\ln 2{\left(x\right)}_0^{\frac{\pi }{2}}$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin 2x\right)dx-\frac{\pi }{2}\ln 2$

Let $t=2x$

$\frac{dt}{2}=dx$

Therefore,

$2I=\frac{1}{2}{\int }_0^{\pi }\ln \left(\sin t\right)dt-\frac{\pi }{2}\ln 2$

$2I=\frac{2}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\sin t\right)dt-\frac{\pi }{2}\ln 2[\because {\int }_0^{a}f\left(x\right)dx=2{\int }_0^{a/2}f\left(x\right)dx]$

$2I={\int }_0^{\frac{\pi }{2}}\ln \left(\sin t\right)dt-\frac{\pi }{2}\ln 2$

$2I=I-\frac{\pi }{2}\ln 2$

$I=-\frac{\pi }{2}\ln 2$

Hence, this is the answer.