Integral help (will use partial fractions method)

In summary: 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
holezch
251
0

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..)
 
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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]
 

1. What is the partial fractions method?

The partial fractions method is a technique used in calculus to simplify a complex rational function into smaller, more manageable parts. It involves breaking down the function into its individual components, each of which can be integrated separately.

2. When should the partial fractions method be used?

The partial fractions method is typically used when integrating rational functions, which are functions that can be expressed as a ratio of polynomials. It is particularly useful when the degree of the numerator is less than the degree of the denominator.

3. How does the partial fractions method work?

The partial fractions method involves expressing a rational function as a sum of simpler fractions, each with a unique denominator. This is achieved by finding the factors of the denominator and setting up a system of equations to solve for the coefficients of each fraction.

4. Are there any limitations to using the partial fractions method?

Yes, the partial fractions method can only be used for rational functions. It also requires the denominator to be factorable into linear and irreducible quadratic terms. Additionally, it may not always be the most efficient method for integrating a function.

5. How can I check my work when using the partial fractions method?

One way to check your work is to substitute the values of the original function into the simplified fractions and make sure they equal the original function. You can also take the derivative of the integrated function to see if it matches the original function.

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