$\int_{-\pi}^\pi\sin^2x\cos^2x\ dx$ with complex integral theorems

Question

Using a complex integral theorem, how can I find this? $$\int_{-\pi}^\pi\sin^2x\cos^2x\ dx$$ I'm not really sure how to approach this problem, since a complex method is asked for.

Answer 1

We have $$\sin^2(x)\cos^2(x) = \dfrac{\sin^2(2x)}4$$ Further, we have $$\int_{-\pi}^{\pi}\sin^2(nx)dx = \int_{-\pi}^{\pi}\cos^2(nx)dx = \dfrac12 \int_{-\pi}^{\pi}\left(\sin^2(nx) + \cos^2(nx) \right)dx = \pi$$ Another way is to recognize that $\sin^2(2x) = \dfrac{1-\cos(4x)}2$. Hence, $$\int_{-\pi}^{\pi}\sin^2(nx)dx = \int_{-\pi}^{\pi} \dfrac{1-\cos(4x)}2dx = \pi$$

Hence, your integral is $\dfrac{\pi}4$.

Answer 2

Using complex integral

$$\sin^2(x)\cos^2(x)=\frac{1}{4}\sin^2(2x)=\frac{1}{4}\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)^2$$

If $z=e^{2ix}$, then $$\sin^2(x)\cos^2(x)=-\frac{1}{16}\left(z-\frac{1}{z}\right)^2=-\frac{1}{16 z^2}(z^2-1)^2$$ and $\mathrm d z=2i e^{2ix}\mathrm d x\implies \frac{1}{2i z}\mathrm d z=\mathrm d x.$

To conclude, $$\int_{-\pi}^\pi \sin^2(x)\cos^2(x)\mathrm d x=-\frac{1}{32 i}\int_{\Gamma }\frac{(z^2-1)^2}{z^3}\mathrm d z,$$ where $\Gamma=\{e^{2ix}\mid x\in [-\pi,\pi]\}$.

Be carreful : If $$\tilde \Gamma=\{e^{i\theta}\mid \theta\in [-\pi,\pi]\},$$

Yout integral is going to be $$...=2\left(-\frac{1}{32 i}\int_{\tilde\Gamma}\frac{(z^2-1)^2}{z^3}\mathrm d z\right).$$

Answer 3

To solve (besides the answer of @surb, this is a way without using complex integrals):

$$\mathcal{I}=\int_{-\pi}^\pi\sin^2(x)\cos^2(x)\space\text{d}x$$

Use:

1. $$\sin^2(x)\cos^2(x)=\frac{\sin^2(2x)}{4}$$
2. $$\sin^2(2x)=\frac{1-\cos(4x)}{2}$$

So, we get:

$$\mathcal{I}=\int_{-\pi}^\pi\sin^2(x)\cos^2(x)\space\text{d}x=\frac{1}{8}\left\{\int_{-\pi}^\pi1\space\text{d}x-\int_{-\pi}^\pi\cos(4x)\space\text{d}x\right\}$$

Use:

1. $$\int1\space\text{d}x=x+\text{C}$$
2. Substitute $u=4x$ and $\text{d}u=4\space\text{d}x$: $$\int_{-\pi}^{\pi}\cos(4x)\space\text{d}x=\frac{1}{4}\int_{-4\pi}^{4\pi}\cos(u)\space\text{d}u$$
3. $$\int\cos(u)\space\text{d}u=\sin(u)+\text{C}$$

So, we get:

$$\mathcal{I}=\int_{-\pi}^\pi\sin^2(x)\cos^2(x)\space\text{d}x=\frac{1}{8}\left\{\left[x\right]_{-\pi}^{\pi}-\frac{1}{4}\cdot\left[\sin(u)\right]_{-4\pi}^{4\pi}\right\}=\frac{\pi}{4}$$