Question Number 211880 by Spillover last updated on 23/Sep/24
Answered by Frix last updated on 23/Sep/24
![∫_0 ^(π/2) ((tan x)/(π^2 +4ln^2 tan x))dx =^([t=2ln tan x]) =(1/2)∫_(−∞) ^∞ (e^t /((t^2 +π^2 )(e^t +1)))dt =^([t=−u]) =−(1/2)∫_∞ ^(−∞) (du/((u^2 +π^2 )(e^u +1)))= =(1/2)∫_(−∞) ^∞ (du/((u^2 +π^2 )(e^u +1)))=(1/2)∫_(−∞) ^∞ (dt/((t^2 +π^2 )(e^t +1))) ⇒ 2∫_0 ^(π/2) ((tan x)/(π^2 +4ln^2 tan x))dx= (1/2)∫_(−∞) ^∞ (e^t /((t^2 +π^2 )(e^t +1)))dt+(1/2)∫_(−∞) ^∞ (dt/((t^2 +π^2 )(e^t +1)))= =(1/2)∫_(−∞) ^∞ (dt/(t^2 +π^2 ))=(1/(2π))[tan^(−1) (t/π)]_(−∞) ^(+∞) =(1/2) ⇒ ∫_0 ^(π/2) ((tan x)/(π^2 +4ln^2 tan x))dx=(1/4)](https://www.tinkutara.com/question/Q211891.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{tan}\:{x}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{tan}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{2ln}\:\mathrm{tan}\:{x}\right]} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{e}^{{t}} }{\left({t}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{t}} +\mathrm{1}\right)}{dt}\:\overset{\left[{t}=−{u}\right]} {=} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\infty} {\overset{−\infty} {\int}}\frac{{du}}{\left({u}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{u}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{du}}{\left({u}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{u}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{t}} +\mathrm{1}\right)} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{tan}\:{x}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{tan}\:{x}}{dx}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{e}^{{t}} }{\left({t}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{t}} +\mathrm{1}\right)}{dt}+\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)\left(\mathrm{e}^{{t}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\pi^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}\pi}\left[\mathrm{tan}^{−\mathrm{1}} \:\frac{{t}}{\pi}\right]_{−\infty} ^{+\infty} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{tan}\:{x}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{tan}\:{x}}{dx}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Spillover last updated on 23/Sep/24
$${correct}.{thanks}.{nice}\:{approach} \\ $$
Answered by Spillover last updated on 23/Sep/24
