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Question Number 123599 by aurpeyz last updated on 26/Nov/20

∫tan^5 xdx

$$\int{tan}^{\mathrm{5}} {xdx} \\ $$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

∫tan^2 x tan^3 x dx  =∫tan^3 x sec^2 xdx−∫tan^3 x dx  =((tan^4 x)/4)−∫sec^2 x tanx −∫tanxdx  =((tan^4 x)/4)−((tan^2 x)/2)+log(cosx)+C

$$\int{tan}^{\mathrm{2}} {x}\:{tan}^{\mathrm{3}} {x}\:{dx} \\ $$$$=\int{tan}^{\mathrm{3}} {x}\:{sec}^{\mathrm{2}} {xdx}−\int{tan}^{\mathrm{3}} {x}\:{dx} \\ $$$$=\frac{{tan}^{\mathrm{4}} {x}}{\mathrm{4}}−\int{sec}^{\mathrm{2}} {x}\:{tanx}\:−\int{tanxdx} \\ $$$$=\frac{{tan}^{\mathrm{4}} {x}}{\mathrm{4}}−\frac{{tan}^{\mathrm{2}} {x}}{\mathrm{2}}+{log}\left({cosx}\right)+{C} \\ $$

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