Practicing Success
$\int\limits^{\frac{\pi}{2}}_{0}sin2x log tan\, x \, dx = $ |
$\frac{\pi}{2}$ $1$ $0$ $\pi $ |
$0$ |
The correct answer is Option (3) → $0$ $I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x \log \tan x \, dx$ ...(1) $I=\int\limits^{\frac{\pi}{2}}_{0}(π-2x)\log\tan(\frac{π}{2}-x)dx$ $I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log\cot xdx$ ...(2) eq. (1) + eq. (2) $2I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log\tan x\cot xdx$ $=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log 1dx$ $=\int\limits^{\frac{\pi}{2}}_{0}0dx=0$ $I=0$ |