We are asked to evaluate the definite integrals given in problems 17 and 19. Problem 17 is: $\int_{-a/2}^{a/2} \cos(\frac{\pi x}{a}) dx$. Problem 19 is: $\int_{0}^{\pi} \sin^2 x dx$.

1. Problem Description

We are asked to evaluate the definite integrals given in problems 17 and

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9. Problem 17 is: $\int_{-a/2}^{a/2} \cos(\frac{\pi x}{a}) dx$.

Problem 19 is: $\int_{0}^{\pi} \sin^2 x dx$.

2. Solution Steps

Problem 17:

We evaluate the integral $\int_{-a/2}^{a/2} \cos(\frac{\pi x}{a}) dx$. Let $u = \frac{\pi x}{a}$. Then $du = \frac{\pi}{a} dx$, so $dx = \frac{a}{\pi} du$. When $x = -a/2$, $u = \frac{\pi}{a} \cdot (-\frac{a}{2}) = -\frac{\pi}{2}$. When $x = a/2$, $u = \frac{\pi}{a} \cdot \frac{a}{2} = \frac{\pi}{2}$.

The integral becomes:

$\int_{-\pi/2}^{\pi/2} \cos(u) \frac{a}{\pi} du = \frac{a}{\pi} \int_{-\pi/2}^{\pi/2} \cos(u) du = \frac{a}{\pi} [\sin(u)]_{-\pi/2}^{\pi/2} = \frac{a}{\pi} [\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})] = \frac{a}{\pi} [1 - (-1)] = \frac{a}{\pi} \cdot 2 = \frac{2a}{\pi}$.

Problem 19:

We evaluate the integral $\int_{0}^{\pi} \sin^2 x dx$. We can use the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$.

Then $\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} dx = \frac{1}{2} \int_{0}^{\pi} (1 - \cos(2x)) dx = \frac{1}{2} [x - \frac{1}{2} \sin(2x)]_{0}^{\pi} = \frac{1}{2} [(\pi - \frac{1}{2} \sin(2\pi)) - (0 - \frac{1}{2} \sin(0))] = \frac{1}{2} [(\pi - 0) - (0 - 0)] = \frac{1}{2} \pi = \frac{\pi}{2}$.

3. Final Answer

For problem 17, the answer is $\frac{2a}{\pi}$.

For problem 19, the answer is $\frac{\pi}{2}$.