# $\int_0^\tau5\cos^4(\theta)+3\cos^2(\theta)+1\mathrm{d}\theta$

Problem:

$$\int_0^{\tau}\sqrt2\cos^4(\theta)+3\cos^2(\theta)+7\mathrm{d}\theta\quad$$

Current strategy:

Replace $\cos^{2}(\theta)$ with $\frac{1}{2}\left(1+\cos(2\theta)\right)$

\begin{align}\int_0^\tau\sqrt2\frac{1}{2}&\left(1+\cos(2\theta)\right)^2(\theta)+3\frac{1}{2}\left(1+\cos(2\theta)\right)+7\mathrm{d}\theta =\\&= \int_0^\tau\sqrt2\frac{1}{2}\left(1+2\cos(\theta)+\cos^2(\theta)\right)+3\frac{1}{2}\left(1+\cos(2\theta)\right)+7\mathrm{d}\theta\\&= \int_0^\tau\sqrt2\frac{1}{2}\left(1+2\cos(\theta)+\frac{1}{2}\left(1+\cos(4\theta)\right)\right)+3\frac{1}{2}\left(1+\cos(2\theta)\right)+7\mathrm{d}\theta \end{align} Is there a better strategy?

• use eulers identity/formula to replace trig functions by exponentials. – MrYouMath Apr 4 '16 at 16:54
• You can also use $\cos(x)^4=\cos(x)^2(1-\sin(x)^2)=\cos(x)^2-\cos(x)^2\sin(x)^2$. Then use $1/2\sin(2x)=\cos(x)\sin(x)$ – MrYouMath Apr 4 '16 at 16:56

These integrals may be dispatched rapidly if you use average values. If we let $\langle f(x)\rangle$ denote the average value of $f(x)$, then since they only differ in phase, $$\langle\cos^2\theta\rangle=\langle\sin^2\theta\rangle=\frac{\langle\cos^2\theta\rangle+\langle\sin^2\theta\rangle}2=\frac{\langle\cos^2\theta+\sin^2\theta\rangle}2=\frac{\langle1\rangle}2=\frac12$$ We may use @MrYouMath's hint above to arrive at \begin{align}\langle\cos^4\theta\rangle & =\langle\cos^2\theta(1-\sin^2\theta)\rangle=\langle\cos^2\theta-\frac14\sin^22\theta\rangle\\ & =\langle\cos^2\theta\rangle-\frac14\langle\sin^22\theta\rangle=\frac12-\frac14\cdot\frac12=\frac38\end{align} Then over an integral number of periods, the trig functions take on their average values, so we can use $$\int_a^bf(x)dx=(b-a)\langle f(x)\rangle$$ So the beauty of it is \begin{align}\int_0^{2\pi}\left(\sqrt2\cos^4\theta+3\cos^2\theta+7\right)d\theta & =2\pi\left(\sqrt2\langle\cos^4\theta\rangle+3\langle\cos^2\theta\rangle+7\langle1\rangle\right)\\ & =2\pi\left(\sqrt2\cdot\frac38+3\cdot\frac12+7\cdot1\right)\\ & =\frac{\pi}4(68+3\sqrt2)\end{align} Just like that!

• Indeed, why not use that one integrates over a period to simplify everything... +1. – Did Apr 8 '16 at 6:26

use that $$\cos(x)^4=\frac{1}{8} (4 \cos (2 x)+\cos (4 x)+3)$$ and $$\cos(x)^2=\frac{1}{2} (\cos (2 x)+1)$$

HINT:

$$\int_0^{2\pi}\left(\sqrt{2}\cos^4(\theta)+3\cos^2(\theta)+7\right)\space\text{d}\theta=$$ $$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\int_0^{2\pi}\cos^2(\theta)\space\text{d}\theta+7\int_0^{2\pi}1\space\text{d}\theta=$$ $$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\int_0^{2\pi}\cos^2(\theta)\space\text{d}\theta+7\left[\theta\right]_{0}^{2\pi}=$$

Use:

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

$$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\int_0^{2\pi}\left[\frac{1}{2}+\frac{\cos(2x)}{2}\right]\space\text{d}\theta+7\left[\theta\right]_{0}^{2\pi}=$$ $$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\left[\frac{1}{2}\int_0^{2\pi}1\space\text{d}\theta+\frac{1}{2}\int_0^{2\pi}\cos(2x)\space\text{d}\theta\right]+7\left[\theta\right]_{0}^{2\pi}=$$ $$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\left[\frac{1}{2}\left[\theta\right]_{0}^{2\pi}+\frac{1}{2}\int_0^{2\pi}\cos(2x)\space\text{d}\theta\right]+7\left[\theta\right]_{0}^{2\pi}=$$

Substitute $u=2\theta$ and $\text{d}u=2\space\text{d}\theta$.

This gives a new lower bound $u=2\cdot0=0$ and upper bound $u=2\cdot2\pi=4\pi$:

$$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\left[\frac{1}{2}\left[\theta\right]_{0}^{2\pi}+\frac{1}{4}\int_0^{4\pi}\cos(u)\space\text{d}u\right]+7\left[\theta\right]_{0}^{2\pi}=$$ $$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+3\left[\frac{1}{2}\left[\theta\right]_{0}^{2\pi}+\frac{1}{4}\left[\sin(u)\right]_{0}^{4\pi}\right]+7\left[\theta\right]_{0}^{2\pi}=$$

For $\cos^4(\theta)$ use the reduction formula:

$$\int\cos^m(x)\space\text{d}x=\frac{\sin(x)\cos^{m-1}(x)}{m}+\frac{m-1}{m}\int\cos^{m-2}(x)\space\text{d}x$$

$$\sqrt{2}\int_0^{2\pi}\cos^4(\theta)\space\text{d}\theta+17\pi$$