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I just need to know why is my method wrong: Let $I=\int\sqrt{\tan x} dx$, let $\tan(x)=t^2$ then, \begin{align*}&\sec^2(x)dx=2tdt\\ \implies&(1+\tan^2(x))dx=2tdt\\ \implies &dx=2tdt/(1+t^2)\end{align*}

So \begin{align*}I &= \int t\cdot \frac{2t}{1+t^2} dt\\ &= \int\frac{2t^2}{1+t^2}dt\end{align*} which can be solved easily. Is this method correct?

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    $\begingroup$ Welcome to MSE. Please use MathJax to format your posts. $\endgroup$
    – saulspatz
    May 20 '20 at 14:08
  • $\begingroup$ "which can be solved easily": why don't you proceed ? $\endgroup$
    – user65203
    May 20 '20 at 14:15
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Let's flesh out @QuantumApple's hint. With $\tan x=t^2$,$$\begin{align}\int\sqrt{\tan x}dx&=\int\frac{2t^2dt}{1+t^4}\\&=\int\frac{1}{2}\left(\tfrac{1}{1-\sqrt{2}t+t^{2}}+\tfrac{1}{1+\sqrt{2}t+t^{2}}-\tfrac{1}{\sqrt{2}}\left(\tfrac{\sqrt{2}-2t}{1-\sqrt{2}t+t^{2}}+\tfrac{\sqrt{2}+2t}{1+\sqrt{2}t+t^{2}}\right)\right)dt\\&=\frac{1}{\sqrt{2}}\arctan(\sqrt{2\tan x}-1)+\frac{1}{\sqrt{2}}\arctan(\sqrt{2\tan x}+1)\\&+\frac{1}{2\sqrt{2}}\ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C.\end{align}$$

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  • $\begingroup$ Thank you, I finally got it. $\endgroup$
    – Eyy boss
    May 21 '20 at 10:43
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If $\tan(x) = t^2$ then $1 + \tan^2(x) = 1 + t^4$ (and not $1 + t^2$).

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  • $\begingroup$ Thank you for pointing that out. $\endgroup$
    – Eyy boss
    May 21 '20 at 10:42

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