this one has a long answer:

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

What do i start with first?

need advices or hints

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- Aug 3rd 2007, 07:03 PM^_^Engineer_Adam^_^integral of \sqrt(\tan(x))dx
this one has a long answer:

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

What do i start with first?

need advices or hints - Aug 3rd 2007, 07:17 PMcurvature
- Aug 3rd 2007, 07:57 PMKrizalid
Look at here.

Don't be afraid about the language, just follow the steps.

In the third one, I applied partial fractions.

I hope it helps. - Aug 3rd 2007, 09:55 PM^_^Engineer_Adam^_^
thanks krizalid,

question:

how is $\displaystyle \int \frac{2u^2}{u^4+1}du$ being factored? coz i used quickmath and gave no results in partial fractions.

once again thanks so much for all the help - Aug 3rd 2007, 10:01 PMKrizalid
$\displaystyle u^4+1=\left(u^2+\sqrt2u+1\right)\left(u^2-\sqrt2u+1\right)$

Then set $\displaystyle \frac{2u^2}{u^4+1}=\frac{ax+b}{u^2+\sqrt2u+1}+\fra c{cx+d}{u^2-\sqrt2u+1}$

This is a little hard. - Aug 3rd 2007, 11:05 PMcurvature
- Aug 3rd 2007, 11:12 PMcurvature
- Aug 4th 2007, 07:49 AMKrizalid
That's the same result as mine :D:D

- Sep 23rd 2007, 10:16 AMKrizalid
After the brute force applied on the problem, there's another way to solve this (elegant by the way).

Suppose it remains to compute

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

Let's use $\displaystyle (x^2+1)+(x^2-1)=2x^2$

Next, we split the integral in two pieces and in each one of them we divide by $\displaystyle x^2$

After some simple calculations we happily get that

$\displaystyle \int\frac{x^2}{x^4+1}\,dx=\frac12\left(\frac1{\sqr t2}\arctan\frac{x-x^{-1}}{\sqrt2}+\frac1{2\sqrt2}\ln\left|\frac{x+x^{-1}-\sqrt2}{x+x^{-1}+\sqrt2}\right|\right)+k$

:D