Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Uncertain about an integral

  1. Feb 10, 2017 #1
    I am trying to evaluate the integral ##\displaystyle \int \frac{x}{1+\cos^2x}dx##. I have started by multiplying both the numerator and the denominator by ##\frac{\sin^2x}{\cos^4x}##, to get ##\displaystyle \int \frac{x\frac{\sin^2x}{\cos^4x}}{1+\tan^2x}dx##, and the denominator simplifies to sec^2, so we get ##\displaystyle \int x \tan^2x~dx##. Is my reasoning up to this point correct?
     
  2. jcsd
  3. Feb 10, 2017 #2

    Mark44

    Staff: Mentor

    I think you have a mistake in your work. If you multiply the denominator by ##\frac{\sin^2x}{\cos^4x}##, you get ##(1+\cos^2x)\frac{\sin^2x}{\cos^4x} = \frac{\sin^2x}{\cos^4x} + \tan^2x##. That first term on the right doesn't simplify to 1.

    Wolframalpha gives a pretty complicated result: http://www.wolframalpha.com/input/?i=integrate+x/(1+++cos^2(x))+dx
     
  4. Feb 12, 2017 #3

    Svein

    User Avatar
    Science Advisor

    Using the complex domain: Substitute [itex]u=e^{ix} [/itex], then [itex] x=\frac{1}{i}\log(u)= -i\log(u)[/itex], [itex] \frac{du}{dx}=ie^{ix}=iu[/itex] and the integrand becomes
    [tex]\frac{-i\log(u)}{1+(\frac{1}{2}(u+\frac{1}{u}))^{2}}=\frac{-i\log(u)}{1+(\frac{1}{4}(u^{2}+2+\frac{1}{u^{2}}))}=\frac{-4iu^{2}\log(u)}{4u^{2}+u^{4}+2u^{2}+1} [/tex]
    This can be rewritten as [tex]\frac{-2iu^{2}\log(u^{2})}{(u^{2}-3-2\sqrt{2})(u^{2}-3+2\sqrt{2})} [/tex] Now substitute z = u2.

    Try to continue from there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Uncertain about an integral
Loading...