0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ use eulers identity/formula to replace trig functions by exponentials. $\endgroup$ – MrYouMath Apr 4 '16 at 16:54
  • 1
    $\begingroup$ 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 $1/2\sin(2x)=\cos(x)\sin(x)$ $\endgroup$ – MrYouMath Apr 4 '16 at 16:56
3
$\begingroup$

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!

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

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

$\endgroup$
1
$\begingroup$

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:

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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.