VITEEE 2015 Paper

Let I = $\int\limits_{0}^{\frac{\pi}{2}}$ (log tan x) . sin 2 x dx ... (i) I = $\int\limits_{0}^{\frac{\pi}{2}}$ log tan $\left(\frac{\pi}{2}-x\right)$ sin 2 $\left(\frac{\pi}{2}-x\right)$ dx [Since $\int\limits_{0}^{a}$ f (x) dx = $\int\limits_{0}^{a}$ f (a - x) dx] ⇒ I = $\int\limits_{0}^{\frac{\pi}{2}}$ log cot x. sin 2x dx ..(ii) [Since sin (π - 2x) = sin 2x] On adding eqs (i) and (ii), we get 2I $\int\limits_{0}^{\frac{\pi}{2}}$ log tan x sin 2 x dx + $\int\limits_{0}^{\frac{\pi}{2}}$ log cot x sin 2x dx = $\int\limits_{0}^{\frac{\pi}{2}}$ sin 2x log (tan x . cot x) dx [Since log m + log n = log (m . n)] = $\int\limits_{0}^{\frac{\pi}{2}}$ sin 2x log dx ⇒ IO = 0 [Since log 1 = 0] ∴ $\int\limits_{0}^{\frac{\pi}{2}}$ sin 2x log (tanx) dx = 0