Question Number 132470 by physicstutes last updated on 14/Feb/21
$$\int\:\mathrm{tan}^{\mathrm{3}} {x}\:{dx} \\ $$
Answered by mindispower last updated on 14/Feb/21
$$\int{tg}\left({x}\right)\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right){dx}−\int{tg}\left({x}\right){dx} \\ $$$$=\frac{{tg}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}+{ln}\mid{cos}\left({x}\right)\mid+{c} \\ $$
Answered by bemath last updated on 14/Feb/21
$$\int\:\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}\:\mathrm{dx}\: \\ $$$$\mathrm{let}\:\mathrm{cos}\:\mathrm{x}\:=\:\mu\: \\ $$$$\int\:\frac{\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\left(−\mathrm{d}\mu\right)}{\mu^{\mathrm{3}} }\:=\:\int\:\left(\frac{\mathrm{1}}{\mu}−\mu^{−\mathrm{3}} \right)\mathrm{d}\mu \\ $$$$=\:\mathrm{ln}\:\mid\mu\mid+\frac{\mathrm{1}}{\mathrm{2}\mu^{\mathrm{2}} }\:+\:\mathrm{c} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{x}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{c} \\ $$