Question

$\int_{0}^{\pi / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

• A. 0
• B. 2
• C. 4
• D. 7

Answer 1

Answer: (A)

$
\text { Let } I=\int_{0}^{\pi / 2}(\log \tan x) \cdot \sin 2 x d x \ldots(i)
$
Let $I=\int_{0}^{\pi / 2}(\log \tan x) \cdot \sin 2 x d x \ldots(i)$ $I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) \sin 2\left(\frac{\pi}{2}-x\right) d x$ ${\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] }$ $I=\int_{0}^{\pi / 2} \log \cot x \cdot \sin 2 x d x \ldots(i i)$ ${[\because \sin (\pi-2 x)=\sin 2 x] }$ On adding eqs $(i)$ and $(i i)$, we get $2 I \int_{0}^{\pi / 2} \log \tan x \cdot \sin 2 x d x+\int_{0}^{\pi / 2} \log \cot x \sin 2 x d x$ $=\int_{0}^{\pi / 2} \sin 2 \log (\tan \cdot \cot x) d x$ ${[\because \log m+\log n=\log (m \cdot n)] }$ $=\int_{0}^{\pi / 2} \sin 2 x \log 1 d x$ $\Rightarrow I=0[\because \log 1=0]$ $\therefore \int_{0}^{\pi / 2} \sin 2 x \log (\tan x) d x=0$