Plz tell me how to integrate this

**Beenish** · January 1st 2010, 02:53 AM

integrate sqrt(tan x)
I substituted tan x = u^2
then i got x = arc tan u^2
By substituting i got 2u^2/(1+u^4)
now what????

**mr fantastic** · January 1st 2010, 03:12 AM

Originally Posted by Beenish

integrate sqrt(tan x)
I substituted tan x = u^2
then i got x = arc tan u^2
By substituting i got 2u^2/(1+u^4)
now what????

Now use partial fractions. Note that

$u^4 + 1 = u^4 - 2u^2 + 2u^2 + 1 = (u^4 - 2u^2 + 1) - 2 u^2$

$= (u^2 - 1)^2 - 2u^2 = (u^2 - 1 - \sqrt{2} u) (u^2 - 1 + \sqrt{2} u)$

$(u^2 - \sqrt{2} u - 1) (u^2 + \sqrt{2} u - 1)$

and these two quadratics are irreducible. The ensuing algebra is your pleasant job to do.

Of interest: integrate Sqrt[Tan[x]] - Wolfram|Alpha

(Be sure to click on Show steps).

**simplependulum** · January 2nd 2010, 11:17 PM

$\int \frac{ x^2}{x^4 + 1} ~dx$

$= \frac{1}{2} \int \frac{ (x^2 - 1) + ( x^2 + 1)}{ x^4 + 1}~dx$

$= \frac{1}{2} \int \frac{ x^2 - 1 }{x^4 + 1} ~dx + \frac{1}{2} \int \frac{ x^2 + 1}{x^4 + 1}~dx$

$= \frac{1}{2} \int \frac{ 1 - \frac{1}{x^2} }{x^2 + x^{-2}}~dx + \frac{1}{2} \int \frac{ 1 + \frac{1}{x^2} }{x^2 + x^{-2}}~dx$

$= \frac{1}{2}\int \frac{d ( x + \frac{1}{x} )}{ (x + \frac{1}{x} )^2 - 2 } + \frac{1}{2}\int \frac{d ( x - \frac{1}{x} )}{ (x - \frac{1}{x} )^2 + 2 }$

Recall the formulae

$\int \frac{du}{u^2 + a^2 } = \frac{1}{a} \tan^{-1}(\frac{u}{a})$

$\int \frac{du}{u^2 - a^2 } = \frac{1}{a} \tanh^{-1}(\frac{u}{a})$

$\frac{1}{2}\int \frac{d ( x + \frac{1}{x} )}{ (x + \frac{1}{x} )^2 - 2 } + \frac{1}{2}\int \frac{d ( x - \frac{1}{x} )}{ (x - \frac{1}{x} )^2 + 2 }$

$= \frac{1}{2\sqrt{2}}\tanh^{-1}(\frac{ x + \frac{1}{x} }{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\tan^{-1}(\frac{ x - \frac{1}{x} }{\sqrt{2}}) + C$

$= \frac{1}{2\sqrt{2}}\tanh^{-1}(\frac{ x^2 + 1 }{\sqrt{2}x}) + \frac{1}{2\sqrt{2}}\tan^{-1}(\frac{ x ^2 - 1}{\sqrt{2}x}) + C$

**Beenish** · January 3rd 2010, 10:25 PM

Thanx a lot. It was very helpful answer.

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