Integral help (will use partial fractions method)

And these integrals can be done by substitution.In summary, the conversation revolved around finding a solution for the integral of sqrt(tanx). The attempt at a solution involved using the substitution u = sqrt(tanx) and factoring the expression into four complex linear factors. The conversation then delved into using the partial fractions method, but it was determined that it would not work in this scenario. The problem was eventually solved using integration by parts.
  • #1

Homework Statement

[tex] \int \sqrt{tanx} dx[/tex]

The Attempt at a Solution

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


[tex] 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)} [/tex]

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..)
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  • #2
Hi holezch! :smile:

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?
  • #3
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 :)
  • #4
Your very last expression can actually be split into the following:
[tex] 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. [/tex]

You need to make more algebraic manipulations after this, but I'm pretty sure this works.
  • #5
thanks, I figured it out.

I solved
[tex] A(u^{2} +\sqrt{2}u + 1) + B(u^{2} -\sqrt{2}u + 1) = 2u^{2},

(A+B)u^{2} + (A - B)\sqrt{2}u + (A+B) = 2u^{2} [/tex]

then, we know that A + B = 0, so A = -B, then [tex] (A+B)u^{2} = 0 [/tex]
and we're left with
[tex] (A-B)\sqrt{2}u = 2u^2,

A-B = \sqrt{2}u,

A = \sqrt{2}/2 u [/tex]

now I have

[tex] 2 \int \frac{ \sqrt{2}/2 u }{(u^2 +\sqrt{2}u + 1)} - \frac{\sqrt{2}/2 u }{(u^2 -\sqrt{2}u + 1)} [/tex]

can I use the partial fractions method now? thanks!
  • #6
okay, I finished it off with parts with each individual integral.. I'm still curious about the partial fraction stuff though :S
  • #7
An alternative solution now that you've done the problem on your own:

[tex] \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[/tex].

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

[tex] \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}[/tex]

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