Evaluate the integral: $\int_{0}^{\pi/2} 3 \sin^2(x) \cos^2 \, dx$

Neetesh Kumar | December 8, 2024

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This is the solution to Math 132
Assignment: 7.2 Question Number 7
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Step-by-step solution:

We aim to evaluate the definite integral $\int_{0}^{\pi/2} 3 \sin^2(x) \cos^2(x) \, dx$.

Step 1: Factor out the constant

The constant $3$ can be factored out of the integral:

$\int_{0}^{\pi/2} 3 \sin^2(x) \cos^2(x) \, dx = 3 \int_{0}^{\pi/2} \sin^2(x) \cos^2(x) \, dx$

Step 2: Use the double-angle identity

To simplify $\sin^2(x) \cos^2(x)$, use the double-angle identity:

$\sin^2(x) \cos^2(x) = \frac{1}{4} \sin^2(2x)$

Substitute this into the integral:

$3 \int_{0}^{\pi/2} \sin^2(x) \cos^2(x) \, dx = 3 \int_{0}^{\pi/2} \frac{1}{4} \sin^2(2x) \, dx$

Simplify:

$3 \int_{0}^{\pi/2} \sin^2(x) \cos^2(x) \, dx = \frac{3}{4} \int_{0}^{\pi/2} \sin^2(2x) \, dx$

Step 3: Use the power-reduction formula

To simplify $\sin^2(2x)$, use the power-reduction identity:

$\sin^2(2x) = \frac{1 - \cos(4x)}{2}$

Substitute this into the integral:

$\frac{3}{4} \int_{0}^{\pi/2} \sin^2(2x) \, dx = \frac{3}{4} \int_{0}^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx$

Simplify:

$\frac{3}{4} \int_{0}^{\pi/2} \sin^2(2x) \, dx = \frac{3}{8} \int_{0}^{\pi/2} \left( 1 - \cos(4x) \right) \, dx$

Step 4: Separate the integral

Distribute the integral:

$\frac{3}{8} \int_{0}^{\pi/2} \left( 1 - \cos(4x) \right) \, dx = \frac{3}{8} \left( \int_{0}^{\pi/2} 1 \, dx - \int_{0}^{\pi/2} \cos(4x) \, dx \right)$

Step 5: Evaluate each term

1. For $\int_{0}^{\pi/2} 1 \, dx$:

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

2. For $\int_{0}^{\pi/2} \cos(4x) \, dx$:

   The integral of $\cos(4x)$ is $\frac{\sin(4x)}{4}$.

Evaluate:

$\int_{0}^{\pi/2} \cos(4x) \, dx = \left[ \frac{\sin(4x)}{4} \right]_{0}^{\pi/2} = \frac{\sin(2\pi)}{4} - \frac{\sin(0)}{4} = 0 - 0 = 0$

Step 6: Combine results

Substitute back into the expression:

$\frac{3}{8} \left( \int_{0}^{\pi/2} 1 \, dx - \int_{0}^{\pi/2} \cos(4x) \, dx \right) = \frac{3}{8} \left( \pi/2 - 0 \right) = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$

Final Answer:

$\int_{0}^{\pi/2} 3 \sin^2(x) \cos^2(x) \, dx = \boxed{\frac{3\pi}{16}}$

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