Solve This Integral with Help from Experts | Forum Discussion

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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.

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I have not worked this out but the substitution u = cos(x) followed by partial fractions seems like it could do the trick.
 
does t=tan(x/2) work in this case?
 
What's wrong with u = cos(x) as suggested above?
 
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"
 
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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)}
 
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.
 
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)
 
Try multiplying the numerator and denominator by [sec(x)]^2, then do the substitutions.
 
  • #10
You might need to substitute twice, meaning you'll want to back-substitute twice into your final answer.
 
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