We have an answer to "How to calculate following integration? ∫(tan⁡x+cot⁡x)dx"

Helen Lewis

Answered question

2022-01-05

How to calculate following integration?

\int (\sqrt{\tan x} + \sqrt{\cot x}) d x

Answer & Explanation

kaluitagf

Beginner2022-01-06Added 38 answers

I = \int (\sqrt{\tan x} + \sqrt{\cot x}) d x

= \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x

Putting

\sin x - \cos x = u, d u = (\cos x + \sin x) d x, u^{2} = 1 - 2 \sin x \cos x, \sin x \cos x = \frac{u^{2} - 1}{2}

I = \int \frac{\sqrt{2} d u}{\sqrt{1 - u^{2}}} = \sqrt{2} \arcsin u + C = \sqrt{2} \arcsin (\sin x - \cos x) + C

where C is an arbitrary constant for indefinite integral.

alexandrebaud43

Beginner2022-01-07Added 36 answers

Substiute

\tan (x) = u = e^{2 t}

\int (\sqrt{\tan (x)} + \sqrt{\cot (x)}) d x = \int (u^{\frac{1}{2}} + u^{- \frac{1}{2}}) \frac{d u}{1 + u^{2}}

= \int \frac{u^{\frac{1}{2}} + u^{- \frac{1}{2}}}{u + u^{- 1}} \frac{d u}{u}

= \int \frac{cosh (t)}{cosh (2 t)} 2 d t

= \int \frac{2}{cosh (2 t)} d sinh (t)

= \int \frac{\sqrt{2}}{1 + 2 {sinh}^{2} (t)} d \sqrt{2} sinh (t)

= \sqrt{2} \arctan (\sqrt{2} sinh (t)) + C

= \sqrt{2} \arctan (\frac{\sqrt{tanh (x)} - \sqrt{\cot (x)}}{\sqrt{2}}) + C

karton

Expert2022-01-11Added 613 answers

Substitute $\tan x = t^{2}$
$\int \sqrt{\tan x} + \sqrt{\cot x} d x = 2 \int \frac{t^{2} + 1}{t^{4} + 1} d t = 2 \int \frac{1 + \frac{1}{t^{2}}}{(t - \frac{1}{t})^{2} + 2} d t$
Now make a $t - \frac{1}{t}$ substitution and we get the answer.

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