Question

Use the symmetry of the graphs of the sine and cosine functions as an aid in evaluating each of the integrals. (a) \(\int_{-\pi / 4}^{\pi / 4} \sin x d x\) (b) \(\int_{-\pi / 4}^{\pi / 4} \cos x d x\) (c) \(\int_{-\pi / 2}^{\pi / 2} \cos x d x\) (d) \(\int_{-\pi / 2}^{\pi / 2} \sin x \cos x d x\)

Short answer

(a) 0 \n (b) \(\sqrt{2}\) \n (c) 2 \n (d) 0

Step-by-step solution

Evaluate the integral of sin(x)

The function \( \sin(x) \) is odd and the limits of integration \(-\pi/4\) and \(\pi/4\) are symmetric about the origin. Using the property of symmetry of odd functions, we can say that the integral from \(-a\) to \(a\) of any odd function is zero. Hence \( \int_{-\pi / 4}^{\pi / 4} \sin x d x = 0 \)

Evaluate the integral of cos(x)

The function \( \cos(x) \) is even and the limits of integration \(-\pi/4\) and \(\pi/4\) are symmetric about the origin. Using the property of symmetry of even functions, we find the integral from 0 to \(a\) and double the result, i.e., \( \int_{-\pi / 4}^{\pi / 4} \cos x d x = 2 \int_{0}^{\pi / 4} \cos x d x \). Evaluating the integral we get \(2 \times (\sin(\pi/4)-\sin(0)) = \sqrt{2}\)

Evaluate the integral of cos(x) from -π/2 to π/2

Again, \( \cos(x) \) is even. The procedure is similar as in Step 2. The integral \( \int_{-\pi / 2}^{\pi / 2} \cos x d x = 2 \int_{0}^{\pi / 2} \cos x d x = 2 \times ( \sin(\pi/2) - \sin(0) ) = 2\)

Evaluate the integral of sin(x)cos(x)

The product \( \sin(x) \cos(x) \) is odd, as it is the product of an odd function and an even function. Hence, using the property of symmetry for odd functions, the integral \( \int_{-\pi / 2}^{\pi / 2} \sin x \cos x d x = 0 \)