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Question Number 23539 by tapan das last updated on 01/Nov/17

∫tan^6 x dx

$$\int\mathrm{tan}\:^{\mathrm{6}} \mathrm{x}\:\mathrm{dx} \\ $$

Answered by $@ty@m last updated on 01/Nov/17

∫tan^4 x(sec^2 x−1)dx ∫tan^4 xsec^2 xdx−∫tan^4 xdxdx =((tan^5 x)/5)−∫tan^2 x(sec^2 x−1)dx =((tan^5 x)/5)−∫tan^2 xsec^2 xdx+∫tan^2 xdx =((tan^5 x)/5)−((tan^3 x)/3)+∫(sec^2 x−1)dx =((tan^5 x)/5)−((tan^3 x)/3)+tan x−x+C

$$\int\mathrm{tan}\:^{\mathrm{4}} {x}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$\int\mathrm{tan}\:^{\mathrm{4}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx}−\int\mathrm{tan}\:^{\mathrm{4}} {xdxdx} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}}{\mathrm{5}}−\int\mathrm{tan}\:^{\mathrm{2}} {x}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}}{\mathrm{5}}−\int\mathrm{tan}\:^{\mathrm{2}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx}+\int\mathrm{tan}\:^{\mathrm{2}} {xdx} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}}{\mathrm{5}}−\frac{\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{3}}+\int\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}}{\mathrm{5}}−\frac{\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{3}}+\mathrm{tan}\:{x}−{x}+{C} \\ $$

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