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This might be a simple question, but how do you evaluate $\int{\sqrt{\tan(x)}\,dx}$ ? I've tried substitution with $u=\tan(x)$, that introduced a $\sec^2(x)$, I've also tried integration by parts, still stuck...

Mårten W
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liya77
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2 Answers2

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Take $u=\sqrt{\tan(x)}$ so that $\sec^2 x = \tan^2 x + 1 = (u^2)^2 + 1 = u^4 + 1$. hence : $$(u^4 + 1) \, dx = 2u \, du $$ Thus: $$dx = \frac{2u \, du}{u^4 + 1}$$

By substitution in the first integral:

$$\int (\tan x)^{1/2} \, dx =\int u \frac{2u \, du}{1 + u^4} =\int \frac{2u^2 \, du}{u^4 + 1}$$ And everything now is a matter of rational function integration, and can be solved using partial fractions. A s a hint for the next steps the denominator can be factored as $(u^2 - \sqrt{2} u + 1)(u^2 + \sqrt{2}u + 1)$

Elaqqad
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First attempt: let $u^2=\tan x$. This gives the integral $$I=\int \frac{2u^2}{u^4+1}\,du$$ which can be integrated by factorising the denominator and using partial fractions. However the factorisation involves surds and is rather messy.

Better: let $u^2=2\tan x$. This gives $$I=\frac1{\sqrt2}\int \frac{4u^2}{u^4+4}\,du =\frac1{\sqrt2}\int \frac{u}{u^2-2u+2}-\frac{u}{u^2+2u+2}\,du\ .$$ Observe that except for the constant out the front, no surds are involved. Now substitute $v=u-1$ for the first bit and $v=u+1$ for the second bit. You will need to be careful with the algebra, but it's not all that bad.

David
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  • Your second method is great, but I wouldn't have known to rewrite the integral $\int{\frac{4u^2}{u^4+4}}du$ as what you did there. I think maybe this requires ingenuity... I just don't know how to come up with these... – liya77 Apr 13 '15 at 01:52
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    You can factorise any polynomial $u^n+a$ by finding the (complex) $n$th roots of $-a$. For $n=4$ there is, alternatively, a nice trick:$$u^4+4=(u^2+2)^2-(2u)^2=(u^2-2u+2)(u^2+2u+2)\ .$$ – David Apr 13 '15 at 01:55