Question

$\displaystyle \sf \int_{0}^{ \frac{\pi}{2} } \frac{ tanx}{4 ln {}^{2} (tanx) + {\pi}^{2} } \: dx∫02π$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \sf \int_{0}^{ \frac{\pi}{2} } \frac{ tanx}{4 ln {}^{2} (tanx) + {\pi}^{2} } \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle \sf \int_{0}^{ \frac{\pi}{2} } \frac{ tanx}{4 ln {}^{2} (tanx) + {\pi}^{2} } \: dx - - - (1)$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm :\longmapsto\:ln(tanx) = y$

$\rm :\longmapsto\:tanx = {e}^{y}$

$\rm :\longmapsto\: {sec}^{2} xdx = {e}^{y} dy$

$\rm :\longmapsto\:dx = \dfrac{{e}^{y}}{ {sec}^{2} x} dy$

$\rm :\longmapsto\:dx = \dfrac{{e}^{y}}{1 + {tan}^{2} x} dy$

$\rm :\longmapsto\:dx = \dfrac{{e}^{y}}{1 + {e}^{2y}} dy$

We know, in definite integrals, when we substitute, we have to change the limits too.

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - \infty \\ \\ \sf \dfrac{\pi}{2} & \sf \infty \end{array}} \\ \end{gathered}$

So, given integral can be rewritten as

$\rm :\longmapsto\:I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{{e}^{y}}{4 {y}^{2} + {\pi}^{2} } \: \times \dfrac{{e}^{y}}{{e}^{2y} + 1} \: dy$

$\rm :\longmapsto\:I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{{e}^{2y}}{(4 {y}^{2} + {\pi}^{2} )({e}^{2y} + 1)} \: dy - - - (2)$

Now, above integral can be also be rewritten as on replacing y by - y as

$\rm :\longmapsto\:I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{{e}^{ - 2y}}{(4 {y}^{2} + {\pi}^{2} )({e}^{ - 2y} + 1)} \: dy$

$\rm :\longmapsto\:I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{1}{(4 {y}^{2} + {\pi}^{2} )({e}^{2y} + 1)} \: dy - - - (3)$

On adding equation (2) and (3), we get

$\rm :\longmapsto\:2I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{1 + {e}^{2y}}{(4 {y}^{2} + {\pi}^{2} )({e}^{2y} + 1)} \: dy$

$\rm :\longmapsto\:2I = \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{1}{4{y}^{2} + {\pi}^{2}} \: dy$

$\rm :\longmapsto\:2I = \dfrac{1}{4} \displaystyle \sf \int_{ - \infty }^{ \infty } \frac{1}{{y}^{2} + \dfrac{ {\pi}^{2} }{4} } \: dy$

$\rm :\longmapsto\:I = \dfrac{1}{8} \times \dfrac{2}{\pi} {tan}^{ - 1}\bigg[\dfrac{2y}{\pi} \bigg]_{ - \infty }^{ \infty }$

$\rm :\longmapsto\:I = \dfrac{1}{8} \times \dfrac{2}{\pi} [{tan}^{ - 1}( \infty ) - {tan}^{ - 1}( - \infty )]$

$\rm :\longmapsto\:I = \dfrac{1}{8} \times \dfrac{2}{\pi} [{tan}^{ - 1}( \infty ) + {tan}^{ - 1}(\infty )]$

$\rm :\longmapsto\:I = \dfrac{1}{8} \times \dfrac{2}{\pi} [2{tan}^{ - 1}( \infty )]$

$\rm :\longmapsto\:I = \dfrac{1}{8} \times \dfrac{2}{\pi} \times 2 \times \dfrac{\pi}{2}$

$\bf\implies \:I = \dfrac{1}{4}$