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Find $$\int \sqrt{\tan x}dx$$

My attempt:

$$\text{Let}\ I=\int \sqrt{\tan(x)}dx$$

$$\text{Let}\ u=\tan(x), du=(1+\tan^{2}(x))dx$$

$$I=\int \frac{\sqrt{u}}{u^{2}+1}$$

$$\text{Let}\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}}$$

$$I=2\int \frac{v^{2}}{v^{4}+1}$$


$$\text{Let}\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}$$

$$\therefore I=\int \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$

$$I=-\int \frac{dt}{1+t^4}$$

Where do I go from here?

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marked as duplicate by no identity, Julian Kuelshammer, Dennis Gulko, Dominic Michaelis, Lord_Farin May 9 '13 at 11:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Could you please tell me how I can know whether there is a duplicate before I post? I will go through the thread and see if the answer I am looking for is there. Thank you. –  please delete me May 9 '13 at 9:57
I changed the question. I do not think it is a duplicate now. –  please delete me May 9 '13 at 10:04
You are asking how to compute the integral in the last line. See accepted answer to that question. It is defenitely a duplicate –  no identity May 9 '13 at 10:12
It is always difficult to determine whether your question have been asked before, just guess how would could look like the most confirmistic variant of your question and use search bar –  no identity May 9 '13 at 10:18

1 Answer 1

First let us compute the following $$\int(\sqrt{\tan x}+\sqrt{\cot x}) dx=\int\frac{\sin x +\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{d(\sin x - \cos x)}{\sqrt{1-(\sin x -\cos x)^2}}$$ which is same as the $$\sqrt 2\int\frac{dz}{\sqrt{1-z^2}}=\sqrt 2\sin ^{-1}z+c$$ Again compute $$\int(\sqrt{\tan x}-\sqrt{\cot x})dx=\int\frac{\sin x -\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{-d(\sin x+\cos x)}{\sqrt{(\sin x+\cos x)^2 -1} } dx$$ and this is same as $$-\sqrt 2\int \frac {dw}{\sqrt{w^2 -1}}=-\sqrt 2\int\frac{\sec u\tan u}{\tan u} du=-\sqrt 2\ln(\sec u +\tan u)+C$$ where $w=\sec u$. Now add both the integrals to obtain the result.

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I think you need to carry this one out. I don't think the minus integral works out as well. –  Ron Gordon May 9 '13 at 10:21

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