Question Number 211727 by Spillover last updated on 18/Sep/24
Answered by MrGaster last updated on 03/Nov/24
![=∫_(−(π/4)) ^(π/4) (((π−4x)tan x(1+tan x))/(1−tan^2 x))dx =∫_(−(π/4)) ^(π/4) (((π−4x)tan x+(π−4x)tan^2 x)/(cos^2 x))dx =∫_(−(π/4)) ^(π/4) (π−4x)tan x sec^2 xdx+∫_(−(π/4)) ^(π/4) (π−4x)tan^2 x sec^2 xdx =∫_(−(π/4)) ^(π/4) (π−4x)tan xd(tan x)+∫_(−(π/4)) ^(π/4) (π−4x)tan^2 xd(tan x) =[(((π−4x)tan^2 x)/2)]_(−(π/4)) ^(π/4) −∫_(−((π )/4)) ^(π/4) (−4)((tan^2 x)/2)dx+[(((π−4x)tan^3 x)/3)]_(−(π/4)) ^(π/4) −∫_(−(π/4)) ^(π/4) (−4)((tan^3 x)/3)dx =[((π tan^2 x)/2)−2 tan^2 x]_(−(π/4)) ^(π/4) +[((π tan^3 x)/3)−((4 tan^3 x)/3)]_(−(π/4)) ^(π/4) =[(π/2)−2]tan^2 (π/4)−[(π/2)−2]tan^2 −(π/4)+[(π/3)−(4/3)]tan^3 (π/4)−[(π/3)−(4/3)]tan^3 −(π/4) =[(π/2)−2]−[(π/2)−2]+[(π/3)−(4/3)]−[(π/3)−(4/3)] =0+0 =0](https://www.tinkutara.com/question/Q213366.png)
$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\pi−\mathrm{4}{x}\right)\mathrm{tan}\:{x}\left(\mathrm{1}+\mathrm{tan}\:{x}\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\pi−\mathrm{4}{x}\right)\mathrm{tan}\:{x}+\left(\pi−\mathrm{4}{x}\right)\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(\pi−\mathrm{4}{x}\right)\mathrm{tan}\:{x}\:\mathrm{sec}^{\mathrm{2}} {xdx}+\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(\pi−\mathrm{4}{x}\right)\mathrm{tan}^{\mathrm{2}} {x}\:\mathrm{sec}^{\mathrm{2}} {xdx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(\pi−\mathrm{4}{x}\right)\mathrm{tan}\:{xd}\left(\mathrm{tan}\:{x}\right)+\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(\pi−\mathrm{4}{x}\right)\mathrm{tan}^{\mathrm{2}} {xd}\left(\mathrm{tan}\:{x}\right) \\ $$$$=\left[\frac{\left(\pi−\mathrm{4}{x}\right)\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{2}}\right]_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} −\int_{−\frac{\pi\:}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(−\mathrm{4}\right)\frac{\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{2}}{dx}+\left[\frac{\left(\pi−\mathrm{4}{x}\right)\mathrm{tan}^{\mathrm{3}} {x}}{\mathrm{3}}\right]_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} −\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left(−\mathrm{4}\right)\frac{\mathrm{tan}^{\mathrm{3}} {x}}{\mathrm{3}}{dx} \\ $$$$=\left[\frac{\pi\:\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{2}}−\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} {x}\right]_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} +\left[\frac{\pi\:\mathrm{tan}^{\mathrm{3}} {x}}{\mathrm{3}}−\frac{\mathrm{4}\:\mathrm{tan}^{\mathrm{3}} {x}}{\mathrm{3}}\right]_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\left[\frac{\pi}{\mathrm{2}}−\mathrm{2}\right]\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}−\left[\frac{\pi}{\mathrm{2}}−\mathrm{2}\right]\mathrm{tan}^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}+\left[\frac{\pi}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\right]\mathrm{tan}^{\mathrm{3}} \frac{\pi}{\mathrm{4}}−\left[\frac{\pi}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\right]\mathrm{tan}^{\mathrm{3}} −\frac{\pi}{\mathrm{4}} \\ $$$$=\left[\frac{\pi}{\mathrm{2}}−\mathrm{2}\right]−\left[\frac{\pi}{\mathrm{2}}−\mathrm{2}\right]+\left[\frac{\pi}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\right]−\left[\frac{\pi}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\right] \\ $$$$=\mathrm{0}+\mathrm{0} \\ $$$$=\mathrm{0} \\ $$
Commented by Spillover last updated on 03/Nov/24
$${great}.{thank}\:{you} \\ $$