Question

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

विकल्प

• π

•  π/2

•  0

• 2π

Answer 1

0 

\[I = \int_0^\frac{\pi}{2} \sin2x \log \tan x\ d x . . . . . \left( 1 \right)\]

\[I = \int_0^\frac{\pi}{2} \sin\left( \pi - 2x \right) \log \tan\left( \frac{\pi}{2} - x \right) d x\]

\[I = \int_0^\frac{\pi}{2} \sin2x \log \cot x\ d x . . . . . \left( 2 \right)\]

\[\text{Adding} \left( 1 \right) and \left( 2 \right), \text{we get}, \]

\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x + \log \cot x \right) d x\]

\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x \cot x  \right) d x\]

\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log1 \right) d x\]

\[I = 0\]