Solve This Integral with Help from Experts | Forum Discussion

[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 couldn't find anything helpful.

The integral is in attachment

 

[Jorriss]

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

 

[Wort]

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

 

[PeroK]

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

 

[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"

 

[Blazejr]

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

 

[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.

 

[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
(which is cos^2(x) which will help you greatly)

 

[phion]

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

 

[phion]

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