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Chapter 7: Problem 28

Evaluate the following integrals. $$\int e^{-2 \theta} \sin 6 \theta d \theta$$

Short Answer

Expert verified
#Question# Evaluate the integral $$\int e^{-2 \theta} \sin 6 \theta d \theta$$ using integration by parts. #Answer# The integral evaluates to $$-\frac{1}{2} e^{-2\theta}\sin 6\theta -\frac{3}{2} e^{-2\theta}\cos(6\theta)+C$$, where C is the integration constant.

Step by step solution

01

Choose u and dv

We have the integral: $$\int e^{-2 \theta} \sin 6 \theta d \theta$$ Let's choose u and dv as follows: $$u = \sin (6 \theta)$$ $$dv = e^{-2 \theta} d \theta$$
02

Find du and v

Now, we need to find du and v. Differentiate u with respect to 胃, and integrate dv with respect to 胃: $$du = 6\cos(6\theta) d\theta$$ $$v = \int e^{-2\theta} d\theta = -\frac{1}{2} e^{-2\theta}$$
03

Apply integration by parts formula

Now, we can apply the integration by parts formula: $$\int e^{-2 \theta} \sin 6 \theta d \theta = uv - \int v du$$ $$= -\frac{1}{2} e^{-2\theta} \sin 6 \theta - \int -\frac{1}{2} e^{-2\theta} (6\cos 6\theta) d \theta$$
04

Integrate the remaining integral

Now, we have another integral to evaluate: $$\int -\frac{1}{2} e^{-2\theta} (6\cos 6\theta) d \theta$$ We can use integration by parts one more time, choosing the appropriate u and dv: $$u = \cos (6 \theta)$$ $$dv = -3 e^{-2 \theta} d \theta$$ Then we differentiate u and integrate dv: $$du = -6\sin(6\theta) d \theta$$ $$v = \frac{3}{2} e^{-2\theta}$$ Using the integration by parts formula again: $$\int -\frac{1}{2} e^{-2\theta} (6\cos 6\theta) d \theta = uv - \int v du$$ $$= -\frac{3}{2} e^{-2\theta} \cos (6 \theta) + \int 9\sin(6\theta)e^{-2\theta} d \theta$$
05

Add the two integrals

Now, we can add the two integrals, which yields: $$-\frac{1}{2} e^{-2\theta}\sin 6\theta -\frac{3}{2} e^{-2\theta}\cos(6\theta)+C$$ where C is the integration constant.

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