Question Number 138009 by bobhans last updated on 09/Apr/21
$$\int\:\mathrm{tan}\:^{\mathrm{3}} \left({x}\right)\:\sqrt{\mathrm{sec}\:^{\mathrm{3}} \left({x}\right)}\:{dx}\:=? \\ $$
Answered by EDWIN88 last updated on 09/Apr/21
$$\mathcal{E}\:=\:\int\:\mathrm{tan}\:^{\mathrm{3}} \left({x}\right)\:\mathrm{sec}\:\left({x}\right)\:\sqrt{\mathrm{sec}\:\left({x}\right)}\:{dx} \\ $$$$=\:\int\:\mathrm{tan}\:\left({x}\right)\mathrm{sec}\:\left({x}\right)\left(\mathrm{sec}\:^{\mathrm{2}} \left({x}\right)−\mathrm{1}\right)\sqrt{\mathrm{sec}\:\left({x}\right)}\:{dx} \\ $$$${letting}\:\ell\:=\sqrt{\mathrm{sec}\:\left({x}\right)}\:\Rightarrow\mathrm{2}\ell\:{d}\ell\:=\:\mathrm{sec}\:\left({x}\right)\mathrm{tan}\:\left({x}\right){dx} \\ $$$$\mathcal{E}\:=\:\int\:\left(\ell^{\mathrm{4}} −\mathrm{1}\right)\ell\:\left(\mathrm{2}\ell\:{d}\ell\:\right) \\ $$$$=\:\mathrm{2}\int\:\left(\ell^{\mathrm{5}} −\ell^{\mathrm{2}} \right){d}\ell \\ $$$$=\mathrm{2}\left(\frac{\ell^{\mathrm{6}} }{\mathrm{6}}−\frac{\ell^{\mathrm{3}} }{\mathrm{3}}\right)+{c}\:=\mathrm{2}\left(\frac{\ell^{\mathrm{6}} }{\mathrm{6}}−\frac{\mathrm{2}\ell^{\mathrm{3}} }{\mathrm{6}}\right)+{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\ell^{\mathrm{3}} \left(\ell^{\mathrm{3}} \:−\mathrm{2}\right)+{c}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{sec}\:^{\mathrm{3}} \left({x}\right)}\:\left(\sqrt{\mathrm{sec}\:^{\mathrm{3}} \left({x}\right)}−\mathrm{2}\right)\:+\:{c}\:\: \\ $$
Answered by liki last updated on 09/Apr/21
Commented by liki last updated on 09/Apr/21
help me pls this question
Commented by mr W last updated on 25/Apr/21
$${you}\:{should}\:{post}\:{your}\:{question}\:{as}\:{new} \\ $$$${question},\:{not}\:{as}\:{answer}\:{or}\:{comment} \\ $$$${to}\:{other}\:{questions}! \\ $$$${read}\:{here}: \\ $$
Commented by mr W last updated on 25/Apr/21
Commented by mr W last updated on 25/Apr/21
$${please}\:{answer}\:{me}\:{if}\:{you}\:{have} \\ $$$${understood}.\:{if}\:{you}\:{don}'{t}\:{follow}\:{the} \\ $$$${rules}\:{of}\:{the}\:{forum},\:{you}\:{may}\:{perhaps} \\ $$$${get}\:{no}\:{help}. \\ $$$${i}'{m}\:{just}\:{trying}\:{to}\:{help}\:{you}. \\ $$