tan-x-cot-x-dx-

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

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

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

\int \frac{\sec^{2} x dx}{\sqrt{1 + \tan^{2} x} . \sqrt{\tan x}} =

\int \frac{du}{\sqrt{u + u^{3}}}, [u = \tan x]

Commented by MJS last updated on 20/May/20

\sqrt{\tan x + \cot x} = \frac{\sqrt{2}}{\sqrt{\sin 2 x}}

this leads to an elliptic integral

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

not elementary calculus sir ?

Answered by MJS last updated on 20/May/20

\int \sqrt{\tan x + \cot x} d x = \sqrt{2} \int \frac{d x}{\sqrt{\sin 2 x}} =

[t = x - \frac{π}{4} \to d x = d t]

= \sqrt{2} \int \frac{d t}{\sqrt{\cos 2 t}} = \sqrt{2} \int \frac{d x}{\sqrt{1 - {2 \sin}^{2} t}} =

= \sqrt{2} F (t ∣ 2) = \sqrt{2} F (x - \frac{π}{4} ∣ 2) + C

search Wikipedia for elliptic integrals

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

thank you prof