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Question Number 112169 by bemath last updated on 06/Sep/20

 ∫ tan^3 x sec^3 x dx ?

$$\:\int\:\mathrm{tan}\:^{\mathrm{3}} {x}\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\:? \\ $$

Answered by Dwaipayan Shikari last updated on 06/Sep/20

∫tanxsecx(sec^2 x)tan^2 x dx  ∫t^2 (t^2 −1)dt             secx=t  (t^5 /5)−(t^3 /3)+C=((sec^5 x)/5)−((sec^3 x)/3)+C

$$\int{tanxsecx}\left({sec}^{\mathrm{2}} {x}\right){tan}^{\mathrm{2}} {x}\:{dx} \\ $$$$\int{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\:{secx}={t} \\ $$$$\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+{C}=\frac{{sec}^{\mathrm{5}} {x}}{\mathrm{5}}−\frac{{sec}^{\mathrm{3}} {x}}{\mathrm{3}}+{C} \\ $$

Answered by bobhans last updated on 06/Sep/20

⇔ ∫ (tan^2  x.sec^2  x) d(sec x) =   ∫ (sec^2 x−1).sec^2 x d(sec x) =  (1/5)sec^5 x−(1/3)sec^3 x + c

$$\Leftrightarrow\:\int\:\left(\mathrm{tan}^{\mathrm{2}} \:\mathrm{x}.\mathrm{sec}^{\mathrm{2}} \:\mathrm{x}\right)\:\mathrm{d}\left(\mathrm{sec}\:\mathrm{x}\right)\:=\: \\ $$$$\int\:\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right).\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{d}\left(\mathrm{sec}\:\mathrm{x}\right)\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sec}\:^{\mathrm{5}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{c} \\ $$

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