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Integral help (will use partial fractions method)

  • Thread starter holezch
  • Start date
  • #1
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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)

so:

[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..)
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
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
251
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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
1,101
3
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
251
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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
251
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okay, I finished it off with parts with each individual integral.. I'm still curious about the partial fraction stuff though :S
 
  • #7
Gib Z
Homework Helper
3,346
5
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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