0 / 2 sin 2 x log tan x ? d x is equal to:

Question

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

Smart Achievers · Accepted Answer

(3)

Ans. 0

Explanation :

$I = \int_{0}^{π / 2} \sin 2 x \log \tan x d x$ $\dots (1)$

$\Rightarrow$ $I = \int_{0}^{π / 2} \sin 2 (\frac{π}{2} - x) \log \tan (\frac{π}{2} - x) d x$

$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$

$= \int_{0}^{π / 2} \sin 2 x \log \cot x d x$ $\dots (2)$

$Adding (1) and (2), we get$

$\Rightarrow$ $2 I = \int_{0}^{π / 2} \sin 2 x [\log \tan x + \log \cot x] d x$

$= \int_{0}^{π / 2} \sin 2 x \cdot \log [\tan x \cdot \cot x] d x$

$= \int_{0}^{π / 2} \sin 2 x \cdot \log 1 d x$ $[∵ \tan x = \frac{1}{\cot x}]$

$∴$ $2 I = 0$ $[∵ \log 1 = 0]$

$\Rightarrow$ $I = 0$

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