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Chapter 7: Problem 6

Use a table of integrals to determine the following indefinite integrals. $$\int \sin 3 x \cos 2 x d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of $$\int \sin 3x \cos 2x dx$$. Answer: $$-\frac{1}{2}\cos(x) - \frac{1}{10}\cos(5x) + C$$

Step by step solution

01

Identify the product-to-sum formula

Using the table of integrals or trigonometric identities, we find the product-to-sum formula for sine and cosine functions with different arguments: $$\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B) + \sin(A+B)]$$ Here, $$A = 3x$$ and $$B = 2x$$.
02

Apply the product-to-sum formula

Plugging the values of A and B into the formula, we get: $$\sin(3x)\cos(2x) = \frac{1}{2}[\sin(3x-2x) + \sin(3x+2x)] = \frac{1}{2}[\sin(x) + \sin(5x)]$$
03

Substitute the product-to-sum expression in the integral

Replace the original expression in the integral with the result from step 2: $$\int \sin 3x \cos 2x dx = \int \frac{1}{2}[\sin(x)+\sin(5x)] dx$$
04

Integrate each term

Integrate each term separately: $$\int \frac{1}{2}[\sin(x)+\sin(5x)] dx = \frac{1}{2}\int \sin(x) dx + \frac{1}{2} \int \sin(5x) dx$$ The integral of $$\sin(x)$$ is $$-\cos(x)$$, and the integral of $$\sin(5x)$$ is $$-\frac{1}{5}\cos(5x)$$.
05

Combine the results and add the constant of integration

Combine the results from the previous step and add the constant of integration, C: $$\frac{1}{2}(-\cos(x)) + \frac{1}{2}\left(-\frac{1}{5}\cos(5x)\right) + C = -\frac{1}{2}\cos(x) - \frac{1}{10}\cos(5x) + C$$ The indefinite integral of $$\int \sin 3x \cos 2x dx$$ is: $$-\frac{1}{2}\cos(x) - \frac{1}{10}\cos(5x) + C$$

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Most popular questions from this chapter

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

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