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Evaluate: $\int \sqrt {\tan (x)} \,dx$

My Attempt: $$=\int \sqrt {\tan (x)} \,dx$$ Let $u=\tan (x)$ $$du=\sec^2 (x) \,dx$$ Then $$=\int \frac {\sqrt {u}}{u^2+1} \,du$$ Let, $s=\sqrt u$ and $ds=\dfrac 1 {2\sqrt u} \, du$ So, $$=2\int \dfrac {s^2}{s^4+1} \, ds$$

What to do further?

Michael Rozenberg
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pi-π
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    Partial fraction decomposition. It might be useful to recall Sophie Germain's identity $$ s^4+4 = (s^2-2s+2)(s^2+2s+2).$$ – Jack D'Aurizio Jan 23 '18 at 16:11
  • If you don't "recall" how to factor $s^4+4,$ you can look at $s^4+4=0,$ so that $s^4 = -4$ and $s^2 = \pm 2i.$ If $s^2 = 2i = 2(\cos90^\circ+i\sin90^\circ),$ then $$s = \pm\sqrt 2 , (\cos45^\circ + i\sin45^\circ) = \pm\sqrt 2, \left( \frac {\sqrt 2} 2 + i \frac{\sqrt 2} 2 \right) = \pm(1+i). $$ Treat $s^2 = -2i$ similarly. Then you have $$ (s-i-1)(s+i-1) = s^2 -2s + 2, $$ and so on. With $s^2+1$ it's a bit messier but the idea is the same. $\qquad$ – Michael Hardy Jan 23 '18 at 16:21

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The rest it's the following. $$\frac{s^2}{s^4+1}=\frac{s^2}{s^4+2s^2+1-2s^2}=\frac{1}{2\sqrt2}\left(\frac{s}{s^2-\sqrt2s+1}-\frac{s}{s^2+\sqrt2s+1}\right)$$ and you get $\ln$ and $\arctan$.

Michael Rozenberg
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