Aug 20, 2014 #1 Wort Messages 4 Reaction score 0 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 Attachments integral.jpg 2.4 KB · Views: 389
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
Aug 20, 2014 #2 Jorriss Messages 1,083 Reaction score 26 I have not worked this out but the substitution u = cos(x) followed by partial fractions seems like it could do the trick.
I have not worked this out but the substitution u = cos(x) followed by partial fractions seems like it could do the trick.
Aug 21, 2014 #4 PeroK Science Advisor Homework Helper Insights Author Gold Member 2024 Award Messages 29,172 Reaction score 20,820 What's wrong with u = cos(x) as suggested above?
Aug 21, 2014 #5 Wort Messages 4 Reaction score 0 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
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"
Aug 21, 2014 #6 Blazejr Messages 23 Reaction score 2 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)}
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)}
Aug 21, 2014 #7 Wort Messages 4 Reaction score 0 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.
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.
Aug 31, 2014 #8 SteliosVas Messages 70 Reaction score 0 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)
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)
Sep 14, 2014 #9 phion Gold Member Messages 175 Reaction score 39 Try multiplying the numerator and denominator by [sec(x)]^2, then do the substitutions.
Sep 14, 2014 #10 phion Gold Member Messages 175 Reaction score 39 You might need to substitute twice, meaning you'll want to back-substitute twice into your final answer.
You might need to substitute twice, meaning you'll want to back-substitute twice into your final answer.