Evaluate the integral: $\int_{0}^{\pi/2} 9 \cos^2(\theta) \, d\theta$

Neetesh Kumar | December 8, 2024

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This is the solution to Math 132
Assignment: 7.2 Question Number 6
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Step-by-step solution:

We aim to evaluate the definite integral $\int_{0}^{\pi/2} 9 \cos^2(\theta) \, d\theta$.

Step 1: Factor out the constant

The constant $9$ can be factored out of the integral:

$\int_{0}^{\pi/2} 9 \cos^2(\theta) \, d\theta = 9 \int_{0}^{\pi/2} \cos^2(\theta) \, d\theta$

Step 2: Use the power-reduction formula

To simplify $\cos^2(\theta)$, we use the power-reduction identity:

$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$

Substitute this into the integral:

$9 \int_{0}^{\pi/2} \cos^2(\theta) \, d\theta = 9 \int_{0}^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta$

Simplify:

$9 \int_{0}^{\pi/2} \cos^2(\theta) \, d\theta = \frac{9}{2} \int_{0}^{\pi/2} \left( 1 + \cos(2\theta) \right) \, d\theta$

Step 3: Separate the integral

Distribute the integral:

$\frac{9}{2} \int_{0}^{\pi/2} \left( 1 + \cos(2\theta) \right) \, d\theta = \frac{9}{2} \left( \int_{0}^{\pi/2} 1 \, d\theta + \int_{0}^{\pi/2} \cos(2\theta) \, d\theta \right)$

Step 4: Evaluate each term

• For $\int_{0}^{\pi/2} 1 \, d\theta$:

$\int_{0}^{\pi/2} 1 \, d\theta = \left[ \theta \right]_{0}^{\pi/2} = \pi/2 - 0 = \pi/2$

• For $\int_{0}^{\pi/2} \cos(2\theta) \, d\theta$:

The integral of $\cos(2\theta)$ is $\frac{\sin(2\theta)}{2}$.

Evaluate:
$\int_{0}^{\pi/2} \cos(2\theta) \, d\theta = \left[ \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0 - 0 = 0$

Step 5: Combine results

Substitute back into the expression:

$\frac{9}{2} \left( \int_{0}^{\pi/2} 1 \, d\theta + \int_{0}^{\pi/2} \cos(2\theta) \, d\theta \right) = \frac{9}{2} \left( \pi/2 + 0 \right) = \frac{9}{2} \cdot \frac{\pi}{2} = \frac{9\pi}{4}$

Final Answer:

$\int_{0}^{\pi/2} 9 \cos^2(\theta) \, d\theta = \boxed{\frac{9\pi}{4}}$

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