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प्रश्न
विकल्प
π
π/2
0
2π
उत्तर
0
\[I = \int_0^\frac{\pi}{2} \sin2x \log \tan x\ d x . . . . . \left( 1 \right)\]
\[I = \int_0^\frac{\pi}{2} \sin\left( \pi - 2x \right) \log \tan\left( \frac{\pi}{2} - x \right) d x\]
\[I = \int_0^\frac{\pi}{2} \sin2x \log \cot x\ d x . . . . . \left( 2 \right)\]
\[\text{Adding} \left( 1 \right) and \left( 2 \right), \text{we get}, \]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x + \log \cot x \right) d x\]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x \cot x \right) d x\]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log1 \right) d x\]
\[I = 0\]
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