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Question Number 131549 by liberty last updated on 06/Feb/21

slowly integral ∫ ((sec^4 x)/( (√(tan^3 x)))) dx =?

$$\mathrm{slowly}\:\mathrm{integral}\: \\ $$$$\int\:\frac{\mathrm{sec}\:^{\mathrm{4}} \mathrm{x}}{\:\sqrt{\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}}}\:\mathrm{dx}\:=? \\ $$

Answered by rs4089 last updated on 06/Feb/21

((2tan^2 x−6)/(3(√(tanx))))+c {c is a constant}

$$\frac{\mathrm{2}{tan}^{\mathrm{2}} {x}−\mathrm{6}}{\mathrm{3}\sqrt{{tanx}}}+{c}\:\:\:\:\left\{{c}\:{is}\:{a}\:{constant}\right\} \\ $$

Answered by EDWIN88 last updated on 06/Feb/21

let (√(tan x)) = u or tan x = u^2 and sec^2 x dx=2u du E=∫(((1+u^4 )(2u du))/u^3 ) = 2∫ (u^2 +u^(−2) ) du E=2((u^3 /3)−(1/u))+ c = (2/3)(√(tan^3 x)) −(2/( (√(tan x)))) + c

$$\:\mathrm{let}\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{u}\:\mathrm{or}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{u}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}=\mathrm{2u}\:\mathrm{du} \\ $$$$\mathrm{E}=\int\frac{\left(\mathrm{1}+\mathrm{u}^{\mathrm{4}} \right)\left(\mathrm{2u}\:\mathrm{du}\right)}{\mathrm{u}^{\mathrm{3}} }\:=\:\mathrm{2}\int\:\left(\mathrm{u}^{\mathrm{2}} +\mathrm{u}^{−\mathrm{2}} \right)\:\mathrm{du} \\ $$$$\mathrm{E}=\mathrm{2}\left(\frac{\mathrm{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{u}}\right)+\:\mathrm{c}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:+\:\mathrm{c} \\ $$

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