# 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.