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- Apply Product-to-Sum Identity: We will use the product-to-sum identities to simplify the integral of the product of two cosines. The identity we will use is:Let's apply this identity to our integral with and .
- Simplify Using Identity: Using the product-to-sum identity, we get:This simplifies to:
- Integrate Each Term: Now we integrate each term separately:(\frac{1}{2})\int \cos(-3x)\,dx + (\frac{1}{2})\int \cos(21x)\,dx\(\newlineSince the integral of \$\cos(kx)\) is \((\frac{1}{k})\sin(kx)\), we can integrate each term.
- Integrate \(\cos(-3x)\): Integrating the first term gives us:\(\newline\)\((\frac{\(1\)}{\(2\)})\int \cos(\(-3\)x)\,dx = (\frac{\(1\)}{\(2\)})(\frac{\(1\)}{(\(-3\))})\sin(\(-3\)x) = -(\frac{\(1\)}{\(6\)})\sin(\(-3\)x)
- Integrate \(\cos(21x)\): Integrating the second term gives us:\(\newline\)\(\frac{\(1\)}{\(2\)}\int\cos(\(21\)x)\,dx = \frac{\(1\)}{\(2\)}(\frac{\(1\)}{\(21\)})\sin(\(21\)x) = \frac{\(1\)}{\(42\)}\sin(\(21\)x)
- Combine Terms: Combining both terms, we get the indefinite integral:\(\newline\)\(\int \cos(9x)\cos(12x)\,dx = -\frac{1}{6}\sin(-3x) + \frac{1}{42}\sin(21x) + C\)
- Simplify First Term: We can simplify the first term by using the property that \(\sin(-\theta) = -\sin(\theta)\):\(-\left(\frac{1}{6}\right)\sin(-3x) = \left(\frac{1}{6}\right)\sin(3x)\) So the final answer is: \(\int \cos(9x)\cos(12x)\,dx = \left(\frac{1}{6}\right)\sin(3x) + \left(\frac{1}{42}\right)\sin(21x) + C\)
