Integral help (will use partial fractions method)

[holezch]

Homework Statement

$$\int \sqrt{tanx} dx$$

The Attempt at a Solution

I used the substitution u = sqrt(tanx), then x = arctan(x^2)

so:

$$2 \int \frac{u^{2}}{u^{4} + 1} du = 2 \int \frac{u^{2}}{(u^{2} + 1)^{2} - 2u^{2}} =2 \int \frac{u^{2}}{(u^2 -\sqrt{2}u + 1)(u^2 +\sqrt{2}u + 1)}$$

now, I suspect that some kind of partial fractions method can be used here, I have 2 irreducible quadratic factors in my denominator, but my numerator has a quadratic term as well.. not sure what to do with that. Please let me know, or give me an idea!

thanks (I'm not looking for a solution, I just need to know how to use the partial fraction method..)

 

[tiny-tim]

Hi holezch!

Partial fractions would only work if you had either u or u2 + 1 (or a linear combination) on the top.

Have you tried factoring it into four (complex) linear factors?

 

[holezch]

doh, it looked so hopeful.. maybe I could somehow make 2u^2 into some that will work? anyway, this is a course on REAL mathematics (I don't think I can use complex factors, or am expected to)

thanks for your reply :)

 

[snipez90]

Your very last expression can actually be split into the following:
$$2 \int \left( -\frac{u}{2\sqrt{2}(-u^2 + u\sqrt{2} -1)} - \frac{u}{2\sqrt{2}(u^2 + u\sqrt{2} + 1)}\right) du.$$

You need to make more algebraic manipulations after this, but I'm pretty sure this works.

 

[holezch]

thanks, I figured it out.

I solved
$$A(u^{2} +\sqrt{2}u + 1) + B(u^{2} -\sqrt{2}u + 1) = 2u^{2}, <br /> <br /> (A+B)u^{2} + (A - B)\sqrt{2}u + (A+B) = 2u^{2}$$

then, we know that A + B = 0, so A = -B, then $$(A+B)u^{2} = 0$$
and we're left with
$$(A-B)\sqrt{2}u = 2u^2,<br /> <br /> A-B = \sqrt{2}u, <br /> <br /> A = \sqrt{2}/2 u$$

now I have

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

can I use the partial fractions method now? thanks!

 

[holezch]

okay, I finished it off with parts with each individual integral.. I'm still curious about the partial fraction stuff though :S

 

[Gib Z]

An alternative solution now that you've done the problem on your own:

$$\int \frac{2u^2}{u^4+1} du = \int \frac{u^2+1}{u^4+1} du + \int \frac{u^2-1}{u^4+1} du$$.

After dividing through numerator and denominator of those integrals by $u^2$ you may notice that it becomes:

$$\int \frac{d\left(u- \frac{1}{u}\right)}{ \left(u- \frac{1}{u}\right)^2 + 2} + \int \frac{d\left(u+ \frac{1}{u}\right)}{ \left(u+ \frac{1}{u}\right)^2 - 2}$$