Problem: ∫ 0 τ 2 cos 4 ⁡ ( θ ) + 3 cos 2 ⁡ ( θ ) + 7 d θ

Question

Problem:      ∫    0          τ            2        cos    4    ⁡  (  θ  )  +  3      cos    2    ⁡  (  θ  )  +  7      d    θ

Asa Austin · Accepted Answer

These integrals may be dispatched rapidly if you use average values. If we let

⟨ f (x) ⟩

denote the average value of

f (x)

, then since they only differ in phase,

⟨ \cos^{2} θ ⟩ = ⟨ \sin^{2} θ ⟩ = \frac{⟨ \cos^{2} θ ⟩ + ⟨ \sin^{2} θ ⟩}{2} = \frac{⟨ \cos^{2} θ + \sin^{2} θ ⟩}{2} = \frac{⟨ 1 ⟩}{2} = \frac{1}{2}

\begin{aligned} ⟨ \cos^{4} θ ⟩ & = ⟨ \cos^{2} θ (1 - \sin^{2} θ) ⟩ = ⟨ \cos^{2} θ - \frac{1}{4} \sin^{2} 2 θ ⟩ \\ = ⟨ \cos^{2} θ ⟩ - \frac{1}{4} ⟨ \sin^{2} 2 θ ⟩ = \frac{1}{2} - \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8} \end{aligned}

Then over an integral number of periods, the trig functions take on their average values, so we can use

\int_{a}^{b} f (x) d x = (b - a) ⟨ f (x) ⟩

So the beauty of it is

\begin{aligned} \int_{0}^{2 π} (\sqrt{2} \cos^{4} θ + 3 \cos^{2} θ + 7) d θ & = 2 π (\sqrt{2} ⟨ \cos^{4} θ ⟩ + 3 ⟨ \cos^{2} θ ⟩ + 7 ⟨ 1 ⟩) \\ = 2 π (\sqrt{2} \cdot \frac{3}{8} + 3 \cdot \frac{1}{2} + 7 \cdot 1) \\ = \frac{π}{4} (68 + 3 \sqrt{2}) \end{aligned}

Just like that!

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