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

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

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

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

Commented by MJS last updated on 20/May/20

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

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

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

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

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

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