Solve an integral with product of trigonometric functions squared

Question

I am trying to solve the following integral. It is possible to integrate by parts but that is way too long so I got another idea.

$$\int _0^{\frac{\pi}{4}}\:\frac{\cos ^2\left(2θ\right)\cos ^2\left(θ\right)}{4}\:d\theta $$

Applying these identities it is supposed to be easier. However, I am stuck. This is what I tried:

(1) $\cos(x)\cos(y)=\frac{1}{2}(\cos(y+x)+\cos(y−x))$

(2) $\cos^2(x)=\frac{1}{2}(\cos(2x)+1)$

Therefore:

$$\frac{1}{4}\int _0^{\frac{\pi }{4}}\cos^2\left(\theta \right)\cos^2\left(2\theta \right)\,d\theta $$ $$\frac{1}{4}\int _0^{\frac{\pi }{4}}\left(\frac{1}{2}\left(\cos\left(2\theta \right)+1\right)\right)\left(\frac{1}{2}\left(\cos\left(4\theta \:\right)\right)+1\right)\,d\theta $$

I tried to use some software to help, and somehow we get to this, that is very easy to integrate: $$\int _0^{\frac{\pi}{4}}\:\frac{\cos\left(6\theta \:\right)+2\cos\left(4\theta \:\right)+3\cos\left(2\theta \:\right)}{8}+\frac{1}{4}\:d\theta \:$$

Answer 1

I think you're very close. The next step is to expand the brackets and then use the 'Product to Sum' formula

$$ \cos(A)\cos(B) = \dfrac{1}{2}\left(\cos(A+B)+\cos(A-B)\right). $$

Answer 2

The general strategy to integrate trigonometric polynomials in $\sin\theta$ and $\cos \theta$ is to linearise them.

An easy method consists in using the definition of these functions in terms of complex exponentials: $$\sin\theta=\frac{\mathrm e^{i\theta}-\mathrm e^{-i\theta}}{2i}, \quad \cos \theta=\frac{\mathrm e^{i\theta}+\mathrm e^{-i\theta}}{2} .$$

To have lighter notations, set $\;u=\mathrm e^{i\theta}$, $\;\bar u=\mathrm e^{-i\theta}$, so that $\sin \theta=\frac1{2i}(u-\bar u)$, $\;\cos\theta=\frac 12(u+\bar u)$. It's simpler to make the calculations with $u$ and $\bar u$, if we take into account the following relations: $$u^n-\bar u^n=2i\sin(n\theta), \qquad u^n+\bar u^n=2\cos(n\theta),\qquad u\,\bar u=1. $$

I'll show how it begins in your case: \begin{align} \bigl(\cos(\theta)\cos(2\theta)\bigr)^2&=\bigl(\tfrac14(u+\bar u)(u^2+\bar u^2)\bigr)^2=\tfrac1{16}\bigl((u^3+u\,\bar u^2+\bar u\,u^2+\bar u^3)\bigr)^2=\tfrac 1{16}\bigl(u^3+\bar u^3+ u+\bar u\bigr)^2\\[1ex] &=\tfrac 1{16}\bigl(u^6+\bar u^6+ u^2+\bar u^2+2(u^3\,\bar u^3+u^4+u^3\,\bar u+\bar u^3u+\bar u^4+u\,\bar u)\bigr)\\[1ex] &=\tfrac 1{16}\bigl(u^6+\bar u^6+2(u^4+\bar u^4)+3(u^2+\bar u^2)+4\bigr) =\dotsm \end{align}