Evaluate the integral. (Remember the constant of integration.) $\int 5 \sin(8x) \cos(5x) \, dx$

Neetesh Kumar | December 7, 2024

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This is the solution to Math 132
Assignment: 7.2 Question Number 4
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Step-by-step solution:

To evaluate the integral $\int 5 \sin(8x) \cos(5x) \, dx$, we proceed step by step.

Step 1: Factor out the constant

The constant $5$ can be factored out of the integral:

$\int 5 \sin(8x) \cos(5x) \, dx = 5 \int \sin(8x) \cos(5x) \, dx$

Step 2: Use the product-to-sum identities

Recall the product-to-sum identity:

$\sin(A) \cos(B) = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right]$

Here, $A = 8x$ and $B = 5x$. Applying the identity:

$\sin(8x) \cos(5x) = \frac{1}{2} \left[ \sin(13x) + \sin(3x) \right]$

Step 3: Rewrite the integral

Substitute back into the integral:

$5 \int \sin(8x) \cos(5x) \, dx = 5 \int \frac{1}{2} \left[ \sin(13x) + \sin(3x) \right] \, dx$

Simplify:

$5 \int \sin(8x) \cos(5x) \, dx = \frac{5}{2} \int \sin(13x) \, dx + \frac{5}{2} \int \sin(3x) \, dx$

Step 4: Integrate each term

The integral of $\sin(kx)$ is:

$\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C$

For $\int \sin(13x) \, dx$:

$\int \sin(13x) \, dx = -\frac{1}{13} \cos(13x)$

For $\int \sin(3x) \, dx$:

$\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x)$

Step 5: Combine results

Substitute these back into the expression:

$\frac{5}{2} \int \sin(13x) \, dx + \frac{5}{2} \int \sin(3x) \, dx = \frac{5}{2} \left(-\frac{1}{13} \cos(13x)\right) + \frac{5}{2} \left(-\frac{1}{3} \cos(3x)\right)$

Simplify:

$\frac{5}{2} \int \sin(13x) \, dx + \frac{5}{2} \int \sin(3x) \, dx = -\frac{5}{26} \cos(13x) - \frac{5}{6} \cos(3x)$

Step 6: Add the constant of integration

Finally, include the constant of integration $C$:

$\int 5 \sin(8x) \cos(5x) \, dx = -\frac{5}{26} \cos(13x) - \frac{5}{6} \cos(3x) + C$

Final Answer:

$\int 5 \sin(8x) \cos(5x) \, dx = \boxed{-\frac{5}{26} \cos(13x) - \frac{5}{6} \cos(3x) + C}$

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