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Question Number 94662 by i jagooll last updated on 20/May/20

∫ (√(tan x+cot x)) dx = ?

$$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$

Commented by i jagooll last updated on 20/May/20

∫ ((sec^2 x dx)/((√(1+tan^2 x)) .(√(tan x)))) =   ∫ (du/(√(u+u^3 )))  , [ u = tan x ]

$$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:.\sqrt{\mathrm{tan}\:\mathrm{x}}}\:=\: \\ $$$$\int\:\frac{\mathrm{du}}{\sqrt{\mathrm{u}+\mathrm{u}^{\mathrm{3}} }}\:\:,\:\left[\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{x}\:\right]\: \\ $$

Commented by MJS last updated on 20/May/20

(√(tan x +cot x))=((√2)/(√(sin 2x)))  this leads to an elliptic integral

$$\sqrt{\mathrm{tan}\:{x}\:+\mathrm{cot}\:{x}}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{sin}\:\mathrm{2}{x}}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{an}\:\mathrm{elliptic}\:\mathrm{integral} \\ $$

Commented by i jagooll last updated on 20/May/20

not elementary calculus sir?

$$\mathrm{not}\:\mathrm{elementary}\:\mathrm{calculus}\:\mathrm{sir}? \\ $$

Answered by MJS last updated on 20/May/20

∫(√(tan x +cot x))dx=(√2)∫(dx/(√(sin 2x)))=       [t=x−(π/4) → dx=dt]  =(√2)∫(dt/(√(cos 2t)))=(√2)∫(dx/(√(1−2sin^2  t)))=  =(√2)F (t∣2) =(√2)F (x−(π/4)∣2) +C  search Wikipedia for elliptic integrals

$$\int\sqrt{\mathrm{tan}\:{x}\:+\mathrm{cot}\:{x}}{dx}=\sqrt{\mathrm{2}}\int\frac{{dx}}{\sqrt{\mathrm{sin}\:\mathrm{2}{x}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}−\frac{\pi}{\mathrm{4}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dt}}{\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}=\sqrt{\mathrm{2}}\int\frac{{dx}}{\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{t}}}= \\ $$$$=\sqrt{\mathrm{2}}\mathrm{F}\:\left({t}\mid\mathrm{2}\right)\:=\sqrt{\mathrm{2}}\mathrm{F}\:\left({x}−\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right)\:+{C} \\ $$$$\mathrm{search}\:\mathrm{Wikipedia}\:\mathrm{for}\:\mathrm{elliptic}\:\mathrm{integrals} \\ $$

Commented by i jagooll last updated on 20/May/20

thank you prof

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$

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