Problem 43 - Calculus Early Transcendentals

Question

Use symmetry to evaluate the following integrals. $$\int_{-\pi / 4}^{\pi / 4} \sec ^{2} x d x$$

Step-by-Step Solution

Answer

Answer: The value of the definite integral is 2.

Step 1: Verify if the function is even or odd

First, we need to verify if the secant function is even or odd. An even function satisfies \(f(x) = f(-x)\), while an odd function satisfies \(f(x) = -f(-x)\). For the secant function, we have $$\sec (-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos x} = \sec x.$$ Thus, the secant function is even. Since the integrand is \(\sec^2(x)\), it is also even.

Step 2: Evaluate the integral of the even function

Since our integrand is even and our interval is symmetric with respect to the origin, we can use the property of even functions to rewrite our integral as: $$\int_{-\pi / 4}^{\pi / 4} \sec ^{2} x d x = 2\int_{0}^{\pi / 4} \sec ^{2} x d x.$$

Step 3: Calculate the definite integral

Now, let's evaluate the definite integral: $$2\int_{0}^{\pi / 4} \sec ^{2} x d x.$$ The integral of \(\sec^2(x)\) is \(\tan(x)\), so we have: $$2(\tan(\pi/4) - \tan(0)).$$

Step 4: Final Solution

Finally, let's calculate the value: $$2(\tan(\pi/4) - \tan(0)) = 2(1 - 0) = 2.$$ So, the value of the integral is \(2\).