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Question Number 167773 by LEKOUMA last updated on 24/Mar/22

Calculate ∫((xtan x)/(cos^4 x))dx

$${Calculate} \\ $$$$\int\frac{{x}\mathrm{tan}\:{x}}{\mathrm{cos}\:^{\mathrm{4}} {x}}{dx} \\ $$

Answered by qaz last updated on 25/Mar/22

I=∫((xtan x)/(cos^4 x))dx =∫x(tan^3 +tan x)d(tan x) =x((1/4)tan^3 x+(1/2)tan x)−∫((1/4)tan^3 x+(1/2)tan x)dx =x((1/4)tan^3 x+(1/2)tan x)−I_1 −I_2 I_1 =(1/4)∫tan^3 xdx=(1/4)∫(sec^2 x−1)tan xdx=(1/8)tan^2 x+(1/4)lncos x+C_1 I_2 =(1/2)∫tan xdx=−(1/2)lncos x+C_2 ⇒I=x((1/4)tan^3 x+(1/2)tan x)−(1/8)tan^2 x+(1/4)lncos x+C

$$\mathrm{I}=\int\frac{\mathrm{xtan}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\mathrm{dx} \\ $$$$=\int\mathrm{x}\left(\mathrm{tan}\:^{\mathrm{3}} +\mathrm{tan}\:\mathrm{x}\right)\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right) \\ $$$$=\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{x}\right)−\int\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{x}\right)\mathrm{dx} \\ $$$$=\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{x}\right)−\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{tan}\:^{\mathrm{3}} \mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)\mathrm{tan}\:\mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{lncos}\:\mathrm{x}+\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{tan}\:\mathrm{xdx}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lncos}\:\mathrm{x}+\mathrm{C}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{I}=\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{x}\right)−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{lncos}\:\mathrm{x}+\mathrm{C} \\ $$

Answered by MJS_new last updated on 25/Mar/22

∫((xtan x)/(cos^4 x))dx=∫((xsin x)/(cos^5 x))dx= u′=((sin x)/(cos^5 x)) → u=(1/(4cos^4 x)) v=x → v′=1 =(x/(4cos^4 x))−(1/4)∫(dx/(cos^4 x))= =(x/(4cos^4 x))−((sin x (1+2cos^2 x))/(12cos^3 x))+C

$$\int\frac{{x}\mathrm{tan}\:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}}{dx}=\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:{u}'=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}}\:\rightarrow\:{u}=\frac{\mathrm{1}}{\mathrm{4cos}^{\mathrm{4}} \:{x}} \\ $$$$\:\:\:\:\:{v}={x}\:\rightarrow\:{v}'=\mathrm{1} \\ $$$$=\frac{{x}}{\mathrm{4cos}^{\mathrm{4}} \:{x}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\mathrm{cos}^{\mathrm{4}} \:{x}}= \\ $$$$=\frac{{x}}{\mathrm{4cos}^{\mathrm{4}} \:{x}}−\frac{\mathrm{sin}\:{x}\:\left(\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{12cos}^{\mathrm{3}} \:{x}}+{C} \\ $$

Commented by peter frank last updated on 26/Mar/22

thanks

$$\mathrm{thanks} \\ $$

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