∫0π2xtan2xln(tanx)dx=

∫0π2xtan2xln(tanx)dxx↦π2−xI=π4∫0π2tan2xln(tanx)dxtanx=sin2x1+cos2xI=π4∫0π2[sin2xcos2xln(sin2x)−sin2xcos2xln(1+cos2x)]dx J=x→π−x−J⇒J=0I=−π8K.K=y=cosx∫−11ln(1+y)ydy

∫0π2xtan2xln(tanx)dx=

∫0π2xtan2xln(tanx)dxx↦π2−xI=π4∫0π2tan2xln(tanx)dxtanx=sin2x1+cos2xI=π4∫0π2[sin2xcos2xln(sin2x)−sin2xcos2xln(1+cos2x)]dx J=x→π−x−J⇒J=0I=−π8K.K=y=cosx∫−11ln(1+y)ydy

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$\int_{0}^{\frac{π}{2}} x \tan 2 x \ln (\tan x) d x =$

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$\frac{π}{2} \int_{0}^{\frac{π}{2}} \tan 2 x \ln (\tan x) d x$

$\frac{π}{4} \int_{0}^{\frac{π}{2}} \tan 2 x \ln (\tan x) d x$

$\frac{π}{8} \int_{- 1}^{1} \frac{\ln (1 - x)}{x} d x$

$\frac{π}{8} \int_{0}^{π} \tan x \ln (1 - \cos x) d x$

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} \int_{0}^{\frac{π}{2}} x \tan 2 x \ln (\tan x) d x \\ x \mapsto \frac{π}{2} - x \\ I = \frac{π}{4} \int_{0}^{\frac{π}{2}} \tan 2 x \ln (\tan x) d x \\ \tan x = \frac{\sin 2 x}{1 + \cos 2 x} \\ I = \frac{π}{4} \int_{0}^{\frac{π}{2}} [\frac{\sin 2 x}{\cos 2 x} \ln (\sin 2 x) - \frac{\sin 2 x}{\cos 2 x} \ln (1 + \cos 2 x)] d x \\ J \overset{x \to π - x}{=} - J \Rightarrow J = 0 \\ I = - \frac{π}{8} K . \\ K \overset{y = \cos x}{=} \int_{- 1}^{1} \frac{\ln (1 + y)}{y} d y \end{array}$