Calculate the integral $\int_{\frac{5\pi}{2}}^{\frac{9\pi}{2}} \cos^4{\theta}\;d\theta.$

Question

Calculate the integral $$\int_{\frac{5\pi}{2}}^{\frac{9\pi}{2}} \cos^4{\theta}\;d\theta.$$

So I know that $$\cos^2{\theta} = \Big(\frac{e^{i\theta} + e^{-i\theta}}{2}\Big)^2 = \frac{1}{4}e^{-2i\theta} + \frac{1}{2} + \frac{1}{4}e^{2i\theta}.$$

But where do I go from here?

Edit: How do I go about solving the integral using Parseval's Theorem?

Answer 1

Hint First, note that the integral is over an interval of length $\require{cancel}2 \pi$ and that the integrand is periodic with period (dividing) $2 \pi$.

Now, squaring the identity you pointed out gives $$\cos^4 \theta = (\cos^2 \theta)^2 = \left[\frac{1}{4}(e^{2 i \theta} + 2 + e^{-2 i \theta})\right]^2 = \frac{3}{8} + \sum a_k e^{k i \theta},$$ for some nonzero integers $k$ and coefficients $a_k$.

For $k \neq 0$ the antiderivative of $e^{k i \theta}$ is $-\frac{i}{k} e^{k i \theta}$, so its integral over an interval $I$ of length $2 \pi$ is zero, and $$\int_I \cos^4 \theta \,d\theta = \int_I \left(\frac{3}{8} + \sum a_k e^{k i \theta}\right) d\theta = \frac{3}{8} \int_I d\theta + a_k \cancelto{0}{\sum \int_I e^{k i \theta} d\theta} = \frac{3}{8} \cdot 2 \pi = \frac{3 \pi}{4} .$$

Answer 2

Hint: Use that $$\cos^4(x)=\frac{1}{8}(3+4\cos(2x)+\cos(4x))$$ For your consideration: the result should be $$\frac{3\pi}{4}$$

Answer 3

Hint:

Choose $y=x-\dfrac{5\pi}2$

to find $$I=\int_0^{2\pi}\sin^4t\ dt$$

By symmetry,

$$2I=\int_0^{\pi/2}(\sin^4t+\cos^4t)dt$$

Now $\sin^4t+\cos^4t=1-2\sin^2t\cos^2t=1-\dfrac{\sin^22t}2=1-\dfrac{1-\cos4t}4$

Answer 4

With Parseval: $$\cos^2(\theta) = \frac{1}{2}+\frac{1}{2}\cos(2\theta) $$ leads to $$ \int_{0}^{2\pi}\left(\cos^2\theta\right)^2\,d\theta = 2\pi\cdot\left(\frac{1}{2}\right)^2 + \pi\cdot\left(\frac{1}{2}\right)^2 = \frac{3\pi}{4}$$ and your integral is the same as the LHS of the last line since $\cos\theta$ is $2\pi$-periodic.

Answer 5

Well, i hesitated to post this, but let it be. Complex analysis. Trigonometry becomes algebra. The integral on the given period interval $J$ of length $2\pi$ of $\cos$ and $\cos^4$, can be rephrased as an integral on the unit circle $C$ centered in the origin, using $\cos t =\frac 12(e^{it}+e^{-it})$. We substitute $z=e^{it}$ and use the residue theorem. $$ \begin{aligned} \int_{J}\cos^4 t\; dt &= \int_{J}\left(\frac 12\left(e^{it}+e^{-it}\right)\right)^4 \; \frac 1{ie^{it}}\; d(e^{it}) \\ &= \int_C \left(\frac 12\left(z+\frac 1z\right)\right)^4 \; \frac 1{iz}\; dz \\ &=2\pi i\cdot\operatorname{Residue}_{z=0} \left(\frac 12\left(z+\frac 1z\right)\right)^4 \; \frac 1{iz} \\ &=2\pi \;\left(\frac 12\right)^4 \binom 42=\frac {3\pi}4\ . \end{aligned} $$ (We had to fish the coefficient of $z$ to the power zero from $\left(z+\frac 1z\right)^4$.) The only advantage of this solution is the immediate generalization, $$ \int_{J}\cos^{2n} t\; dt = 2\pi\; \left(\frac 12\right)^{2n} \binom {2n}n\ . $$ But it is simpler to get it from the simplest solution in this party of answers, the one of Travis. (+1 from me above.)