I have to solve this indefinite integral $$\int_{0}^{\pi} \frac{\cos(4\theta)}{1+\cos^2(\theta)}\, d\theta$$

I changed $\cos(4\theta)$ for $\frac{e^{4i\theta}+e^{-4i\theta}}{2}$ on the unit disk, but my teacher told me that this shouldn't be done.

What should I do?

  • 3
    $\begingroup$ First, this is not an indefinite integral. You want to integrate around the entire circle and write it as a contour integral $\int_{|z|=1} f(z)\,dz$ for an appropriate function $f(z)$. $\endgroup$ – Ted Shifrin May 21 at 0:25
  • $\begingroup$ sorry for my mistake. I did try to find the appropiate $f(z)$ by doing the change that I mentioned $\endgroup$ – Estrellita42 May 21 at 0:28

Use the formula $$\cos 2x=2\cos^2 x-1,$$ you will get $$\cos(4\theta)=8\cos^4 \theta-8\cos^2 \theta+1.$$

Let $t=1+\cos^2(\theta)$, then $$\frac{\cos(4\theta)}{1+\cos^2(\theta)}=8 t-24+\frac{17}{t}.$$

$$\int_{0}^{\pi} \frac{\cos(4\theta)}{1+\cos^2(\theta)}\, d\theta =2\int_{0}^{\pi/2} \frac{\cos(4\theta)}{1+\cos^2(\theta)}\, d\theta$$ $$=2\int_{0}^{\pi/2} \left(8(1+\cos^2 \theta)-24+\frac{17}{1+\cos^2 \theta}\right) \, d\theta$$ $$=\left(-12+\frac{17}{\sqrt 2}\right)\pi.$$


Hint: $$I=\int_{0}^{\pi} \frac{2\cos^22\theta-1}{1+\frac12(1+\cos2\theta)} d\theta$$ with substitution $\phi=2\theta$ we have $$I=\int_{0}^{2\pi} \frac{2\cos^2\phi-1}{3+\cos \phi} d\phi$$ now take $\cos \phi=\dfrac12(z+\dfrac1z)$ and $d\phi=\dfrac{dz}{iz}$ and use residue theorem.

  • $\begingroup$ Isn't this literally the same thing as $z=e^{i\theta}$, which his/her teacher told not to? $\endgroup$ – acarturk May 21 at 0:46
  • $\begingroup$ it is on unit circle. $\endgroup$ – Nosrati May 21 at 0:48
  • $\begingroup$ The point is to end up with a rational function $f(z)$ to integrate around the unit circle by applying the Residue Theorem. You didn't say that explicitly, but it's there. $\endgroup$ – Ted Shifrin May 21 at 0:50

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