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Chapter 5: Problem 53

Integrals with \(\sin ^{2} x\) and \(\cos ^{2} x\) Evaluate the following integrals. $$\int_{-\pi}^{\pi} \cos ^{2} x d x$$

Short Answer

Expert verified
Answer: The value of the integral \(\int_{-\pi}^{\pi} \cos ^{2} x d x\) is \(\pi\).

Step by step solution

01

Use the trigonometric identity for \(\cos^2 x\)

Replace \(\cos^2 x\) with \(\frac{1 + \cos(2x)}{2}\) in the given integral: $$\int_{-\pi}^{\pi} \cos ^{2} x d x = \int_{-\pi}^{\pi} \frac{1 + \cos(2x)}{2} d x$$
02

Separate the integral into two parts

Split the integral into two separate integrals, one for the constant term and another for the trigonometric term: $$\int_{-\pi}^{\pi} \frac{1 + \cos(2x)}{2} d x = \frac{1}{2} \int_{-\pi}^{\pi} d x + \frac{1}{2} \int_{-\pi}^{\pi} \cos(2x) d x$$
03

Evaluate the first integral and the second integral separately

Evaluate each integral. For the first integral, apply the integral of the constant term and apply the fundamental theorem of calculus for the second integral: $$\frac{1}{2} \int_{-\pi}^{\pi} d x = \frac{1}{2}(x\Big|_{-\pi}^{\pi}) = \frac{1}{2} (\pi - (-\pi)) = \pi$$ $$\frac{1}{2} \int_{-\pi}^{\pi} \cos(2x) d x = \frac{1}{2} \left[\frac{1}{2}\sin(2x)\right]_{-\pi}^{\pi} = \frac{1}{4}[\sin(2\pi) - \sin(-2\pi)] = \frac{1}{4}(0 - 0) = 0$$
04

Combine the results

Since the second integral evaluates to 0, the final result of the given integral is equal to the value of the first integral: $$\int_{-\pi}^{\pi} \cos ^{2} x d x = \pi$$

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