Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\frac{\pi }{2}}(2logsinx-logsin2x)dx.$

Answer 1

Let$I={\int }_0^{\frac{\pi }{2}}(2logsinx-logsin2x)dx={\int }_0^{\frac{\pi }{2}}(logsi{n}^{2}x-logsin2x)dx={\int }_0^{\frac{\pi }{2}}\left(\frac{si{n}^{2}x}{sin2x}\right)dx[\because logm-logn=log\frac{m}{n}]={\int }_0^{\frac{\pi }{2}}log\left(\frac{si{n}^{2}x}{2sinxcosx}\right)dx[\because sin2x=2sinxcosx]={\int }_0^{\frac{\pi }{2}}log\left(\frac{tanx}{2}\right)dx={\int }_0^{\frac{\pi }{2}}log\left(tanx\right)-log2dx={\int }_0^{\frac{\pi }{2}}log\left(tanx\right)dx-{\int }_0^{\frac{\pi }{2}}log2dx\Rightarrow I=I_1-\left(log2\right)[x{]}_0^{\frac{\pi }{2}}=I_1-(\frac{\pi }{2}-0)log2$
Where$I_1{\int }_0^{\frac{\pi }{2}}log\left(tanx\right)dx\Rightarrow I_1={\int }_0^{\frac{\pi }{2}}log\left(tan(\frac{\pi }{2}-x)\right)dx[\because I={\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]\Rightarrow I_1={\int }_0^{\frac{\pi }{2}}log\left(cotx\right)dx$
on adding eqs. (ii) and (iii) we get
$2I_1={\int }_0^{\frac{\pi }{2}}\left(log\right(tanx)+log(cotx\left)\right)dx={\int }_0^{\frac{\pi }{2}}log\left(tanxcotx\right)dx[\because logm+logn=log(mn\left)\right]={\int }_0^{\frac{\pi }{2}}log1dx=0\Rightarrow I_1=0(\because tanx=\frac{1}{cotx})$
on putting th evalue of $I_1$ in Eq (i) we get $I=0-\frac{\pi }{2}log2=-\frac{\pi }{2}log2$