tan-x-cot-x-dx-

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

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

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

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

Commented by MJS last updated on 20/May/20

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

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

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

Answered by MJS last updated on 20/May/20

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

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

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

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