How to calculate $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals

Question

How do I go about calculating $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals?

Answer 1

By symmetry, $$ I = \int_{0}^{\pi}\frac{\cos(x)^4}{1+\sin(x)^2}\,dx = 2 \int_{0}^{\pi/2}\frac{\cos(x)^4}{1+\sin(x)^2}\,dx $$ and by substituting $x=\arctan(t)$ we get $$ I = 2 \int_{0}^{+\infty}\frac{dt}{(1+t^2)^2 (1+2t^2)}=\int_{-\infty}^{+\infty}\frac{dt}{(1+t^2)^2(1+2t^2)}. $$ The integrand function is now a meromorphic function with double poles at $t=\pm i$, simple poles at $t=\pm\frac{i}{\sqrt{2}}$, bounded by $\frac{C}{|t|^2}$ as $|t|\to +\infty$. By the ML lemma and the residue theorem it follows that: $$ I = 2\pi i\sum_{\xi\in\{i,i/\sqrt{2}\}}\!\!\!\text{Res}\left(\frac{1}{(1+t^2)^2(1+2t^2)},t=\xi\right)=2\pi i\left(\frac{5i}{4}-i\sqrt{2}\right)=\color{red}{\frac{\pi}{2}(4\sqrt{2}-5)}. $$

Answer 2

Suppose we seek to evaluate $$J = \int_0^{\pi} \frac{\cos^4 x}{1+\sin^2 x} dx = \frac{1}{2} \int_0^{2\pi} \frac{\cos^4 x}{1+\sin^2 x} dx.$$

Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence $\frac{dz}{iz} = dx$ to obtain

$$\frac{1}{2} \int_{|z|=1} \frac{(z+1/z)^4/2^4}{1+(z-1/z)^2/(2i)^2} \frac{dz}{iz} \\ = -\frac{1}{8} \int_{|z|=1} \frac{(z^2+1)^4}{-4z^2+(z^2-1)^2} \frac{dz}{iz^3} \\ = \frac{i}{8} \int_{|z|=1} \frac{(z^2+1)^4}{(z^2-1)^2-4z^2} \frac{z\; dz}{z^4}.$$

Now put $w=z^2$ so that $z\; dz = \frac{1}{2} \; dw$ and note that this loops around the origin twice so that we must multiply a single loop by two to get

$$2\times \frac{i}{8} \int_{|w|=1} \frac{(w+1)^4}{(w-1)^2-4w} \frac{1}{2} \frac{dw}{w^2} \\ = \frac{i}{8} \int_{|w|=1} \frac{(w+1)^4}{w^2-6w+1} \frac{dw}{w^2}.$$

In addition to the double pole at $w=0=\rho$ there are two simple poles here at

$$\rho_{0,1} = 3 \pm 2\sqrt{2}$$

and clearly $\rho_1$ is the only one inside the contour. Converting to partial fractions we get

$$\frac{64}{w^2-6w+1} + 1 + \frac{10}{w} + \frac{1}{w^2}.$$

Hence the non-zero contribution comes from

$$\frac{64}{w^2-6w+1} + \frac{10}{w}.$$

The first term yields

$$\mathrm{Res}_{w=\rho_1} \frac{64}{w^2-6w+1} = \left. \frac{64}{2w-6} \right|_{w=\rho_1} \\ = \frac{64}{-4\times\sqrt{2}} = -\frac{16}{\sqrt{2}} = -8\sqrt{2}.$$

The second term produces $10$ by inspection. Collecting everything we have

$$2\pi i \times \frac{i}{8} \times(10 - 8\sqrt{2}) = \frac{\pi}{4} (8\sqrt{2}-10)$$

or

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{2}(4\sqrt{2}-5).}$$