# integral of \sqrt(\tan(x))dx

• Aug 3rd 2007, 07:03 PM
integral of \sqrt(\tan(x))dx
this one has a long answer:
$\displaystyle \int\sqrt{\tan{x}}dx$

• Aug 3rd 2007, 07:17 PM
curvature
Quote:

this one has a long answer:
$\displaystyle \int\sqrt{\tan{x}}dx$

You may try this substitution:u=sqrt(tanx).
• Aug 3rd 2007, 07:57 PM
Krizalid
Look at here.

In the third one, I applied partial fractions.

I hope it helps.
• Aug 3rd 2007, 09:55 PM
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 PM
Krizalid
$\displaystyle u^4+1=\left(u^2+\sqrt2u+1\right)\left(u^2-\sqrt2u+1\right)$

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 PM
curvature
Quote:

how is $\displaystyle \int \frac{2u^2}{u^4+1}du$ being factored?

At this step you may use the reciprocal subsititution:u=1/t. This may make the integral easier.
• Aug 3rd 2007, 11:12 PM
curvature
Quote:

thanks krizalid,
question:
$\displaystyle \int \frac{2u^2}{u^4+1}du$

Maple gives the following result:
• Aug 4th 2007, 07:49 AM
Krizalid
That's the same result as mine :D:D
• Sep 23rd 2007, 10:16 AM
Krizalid
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