Question

${\int }_0^{\pi /2}\sin 2x\cdot \log \tan xdx$ is equal to

• A. 0
• B. 2
• C. 4
• D. 7

Answer 1

Answer: (A)

$<br/><br/>\text{ Let }I={\int }_0^{\pi /2}(\log \tan x)\cdot \sin 2xdx\dots (i)<br/><br/>$
Let $I={\int }_0^{\pi /2}(\log \tan x)\cdot \sin 2xdx\dots (i)$ $I={\int }_0^{\pi /2}\log \tan \left(\frac{\pi }{2}-x\right)\sin 2\left(\frac{\pi }{2}-x\right)dx$ $\left[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx\right]$ $I={\int }_0^{\pi /2}\log \cot x\cdot \sin 2xdx\dots (ii)$ $[\because \sin (\pi -2x)=\sin 2x]$ On adding eqs $(i)$ and $(ii)$, we get $2I{\int }_0^{\pi /2}\log \tan x\cdot \sin 2xdx+{\int }_0^{\pi /2}\log \cot x\sin 2xdx$ $={\int }_0^{\pi /2}\sin 2\log (\tan \cdot \cot x)dx$ $[\because \log m+\log n=\log (m\cdot n)]$ $={\int }_0^{\pi /2}\sin 2x\log 1dx$ $\Rightarrow I=0[\because \log 1=0]$ $\therefore {\int }_0^{\pi /2}\sin 2x\log (\tan x)dx=0$