Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 35

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Hint:

To solve this equation we use $\int_{a}^{b} f(x) d x$   formula.

Given:

$\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \; d x$

Solution:

Let

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \; d x \ldots(i) \\\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2\left(\frac{\pi}{2}-x\right) \cdot \log \left[\tan \left(\frac{\pi}{2}-x\right)\right] d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \sin (\pi-2 x) \cdot \log \cot x \; d x \end{aligned}

$=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \cot x\; d x \ldots(i i)$

$2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \tan x+\sin 2 x \cdot \log \cot x \; d x$

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x+\log \cot x) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x \cdot \cot x) d x \end{aligned}

\begin{aligned} &\log a \cdot b=\log a+\log b \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x\left(\log \tan x \cdot \frac{1}{\tan x}\right) d x \end{aligned}

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log 1) d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}