I'm looking to solve the following integral using substitution:

$$\int \frac{dx}{2-\cos x}$$

Let $z=\tan\frac{x}{2}$

Then $dz=\frac 1 2 \sec^2 \frac x 2\,dx$

$$\sin x=\frac{2z}{z^2+1}$$

$$\cos x =\frac{1-z^2}{z^2+1}$$


$$\int \frac{dx}{2-\cos x} = \int \frac{\frac{2\,dz}{z^2+1}}{2-\frac{1-z^2}{z^2+1}} =\int \frac{2\,dz}{3z^2+1}$$

But this is where things start to look at bit sticky. If I integrate this last fraction, then I get a very complex expression that seems to defeat the point of z-substitution. Any suggestions for where I may be going wrong?



Thank you for your feedback. I've completed my work as per your suggestions:

$$\int \frac{2\,dz}{3z^2+1} = 2\cdot\left(\frac{\tan^{-1} \frac{z}{\sqrt{3}}}{\sqrt{3}} \right) = \frac{2\tan^{-1} \left(\sqrt{3}\tan{\frac{x}{2}}\right)}{\sqrt{3}}+c$$

  • 1
    $\begingroup$ $2(z^2+1)-(1-z^2)=2z^2+z^2+2-1=3z^2+1$ $\endgroup$
    – randomgirl
    May 19 '15 at 20:01
  • 1
    $\begingroup$ last denominator should be $3z^2+1$ $\endgroup$
    – WW1
    May 19 '15 at 20:01
  • $\begingroup$ Look up arctan(x) derivative $\endgroup$
    – ntarki
    May 19 '15 at 20:02
  • $\begingroup$ Similar problems (and maybe exactly this one, too) have already been answered on MSE many times. Use the Weierstrass half-angle substitution, the residue theorem or both. $\endgroup$ May 19 '15 at 20:10


$$\int \frac{1}{a^2+x^2}dx=\frac1a \arctan(x/a)+C$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.