Need help with this integral

1. Aug 20, 2014

Wort

Hey. I'm new to the forum and I was hoping you could help me solve this integral. I was searching for a clue on similar integrals posted on internet, but I couldnt find anything helpful.

The integral is in attachment

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2. Aug 20, 2014

Jorriss

I have not worked this out but the substitution u = cos(x) followed by partial fractions seems like it could do the trick.

3. Aug 21, 2014

Wort

does t=tan(x/2) work in this case?

4. Aug 21, 2014

PeroK

What's wrong with u = cos(x) as suggested above?

5. Aug 21, 2014

Wort

Nothing really. I just have difficulties recognizing what I should take as a substitution.

EDIT: If I take u=cosx, du=-sinx dx which makes dx=-du/sinx ... and everywhere else there's "u" instead of "cosx"

Last edited: Aug 21, 2014
6. Aug 21, 2014

Blazejr

You already have sin x dx under the integral. That becomes -du, all your cosines become u, and you get integral of $\frac{4(u-1)}{u^2(2-u)}$

7. Aug 21, 2014

Wort

wow...just WOW... I didnt even notice there's "sinx" in the numerator, how stupid of me. Thank you and I'm really sorry for wasting your time.

8. Aug 31, 2014

SteliosVas

If you do integration by substitution than partial fractions you can get the the right answer

You also had sin(x)sin(x) at the bottom

9. Sep 14, 2014

phion

Try multiplying the numerator and denominator by [sec(x)]^2, then do the substitutions.

10. Sep 14, 2014

phion

You might need to substitute twice, meaning you'll want to back-substitute twice into your final answer.