Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\sin 2x\log \tan xdx$ is equal to

(a) π
(b) π/2
(c) 0
(d) 2π

Answer 1

(c) 0

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x\log \tan xdx.....\left(1\right) \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \left(\mathrm{\pi }-2x\right)\log \tan \left(\frac{\mathrm{\pi }}{2}-x\right)dx \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x\log \cot xdx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}, \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x\left(\log \tan x+\log \cot x\right)dx \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x\left(\log \tan x\cot x\right)dx \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2x\left(\log 1\right)dx \\ I=0 \end{gathered}$$