Question Number 125889 by bramlexs22 last updated on 15/Dec/20
$$\:\:\int\:\frac{\mathrm{tan}\:^{\mathrm{3}} {x}}{\:\sqrt{\mathrm{sec}\:{x}}}\:{dx}\:?\: \\ $$
Answered by bobhans last updated on 15/Dec/20
![∫ ((tan x(sec^2 x−1))/( (√(sec x)))) dx = [ sec x = u^2 ⇒tan x dx = ((2 du)/u) ] I=∫ (((u^4 −1))/u)(((2du)/u))=2∫(u^2 −u^(−2) )du = 2((1/3)u^3 +(1/u))+c = ((2(√(sec^3 x)))/3) + (2/( (√(sec x)))) + c](Q125891.png)
$$\int\:\frac{\mathrm{tan}\:{x}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{sec}\:{x}}}\:{dx}\:= \\ $$$$\:\left[\:\mathrm{sec}\:{x}\:=\:{u}^{\mathrm{2}} \:\Rightarrow\mathrm{tan}\:{x}\:{dx}\:=\:\frac{\mathrm{2}\:{du}}{{u}}\:\right] \\ $$$${I}=\int\:\frac{\left({u}^{\mathrm{4}} −\mathrm{1}\right)}{{u}}\left(\frac{\mathrm{2}{du}}{{u}}\right)=\mathrm{2}\int\left({u}^{\mathrm{2}} −{u}^{−\mathrm{2}} \right){du} \\ $$$$\:=\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} +\frac{\mathrm{1}}{{u}}\right)+{c}\: \\ $$$$\:=\:\frac{\mathrm{2}\sqrt{\mathrm{sec}\:^{\mathrm{3}} {x}}}{\mathrm{3}}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{sec}\:{x}}}\:+\:{c}\: \\ $$
Answered by Dwaipayan Shikari last updated on 15/Dec/20
$$\int\frac{{tanxsec}^{\mathrm{2}} {x}}{\:\sqrt{{secx}}}−\frac{{tanx}}{\:\sqrt{{secx}}}{dx}\:\:\:\:\:\:\:\:\:\:\:\:{secx}={t}^{\mathrm{2}} \Rightarrow{secx}\:{tanx}=\frac{{dt}}{{dx}} \\ $$$$=\mathrm{2}\int{t}^{\mathrm{2}} {dt}\:−\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{\mathrm{2}}{{t}}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{sec}^{\mathrm{3}} {x}}\:+\frac{\mathrm{2}}{\:\sqrt{{secx}}}+{C} \\ $$