Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises: 17

Answer

$$\displaystyle\int e^{2\theta}\sin{3\theta}d\theta= \frac{2}{13}e^{2\theta}\sin{3\theta}-\frac{3}{13}e^{2\theta}\cos{3\theta}+C$$

Work Step by Step

We use integration by parts to obtain: $u= \sin{3\theta}\quad du=3cos{3\theta}d\theta$ $dv = e^{2\theta}d\theta\quad v=\frac{1}{2}e^{2\theta}$ $\int u\thinspace dv = uv - \int v\thinspace du$ $$\int e^{2\theta}\sin{3\theta}=\sin{3\theta}\frac{1}{2}e^{2\theta}-\int \frac{1}{2} e^{2\theta}3\cos{3\theta}\thinspace d\theta$$ $$\int e^{2\theta}\sin{3\theta}=\sin{3\theta}\frac{1}{2}\thinspace e^{2\theta}-\frac{3}{2}\int e^{2\theta}\cos{3\theta}\thinspace d\theta$$ $u=\cos{3\theta}\quad du = -3\sin{3\theta}$ $dv=e^{2}{d\theta}\quad v=\frac{1}{2}e^{2\theta}$ $$\int e^{2\theta}\sin{3\theta}\thinspace \frac{1}{2}e^{2\theta}-\frac{3}{2}(uv-\int v\thinspace du)$$ $$= \sin{3\theta}\thinspace\frac{1}{2}e^{2\theta}-\frac{3}{2}\bigg(\cos({3\theta})\thinspace\frac{1}{2}e^{2\theta}-\int\frac{1}{2}e^{2\theta}\times -3\sin{3\theta}\thinspace d\theta\bigg)$$ $$=\sin{3\theta}\times\frac{1}{2}e^{2\theta}-\frac{3}{2}\bigg(\frac{1}{2}e^{2\theta}\cos{3\theta}+\frac{3}{2}\int e^{2\theta}\sin{3\theta}\thinspace d\theta\bigg)$$ $$=\frac{1}{2}e^{2\theta}\sin{3\theta}-\frac{3}{4}e^{2\theta}\cos{3\theta}-\frac{9}{4}\int e^{2\theta}\sin{3\theta}\thinspace d\theta$$ $$\frac{13}{4}\int e^{2\theta}\sin{3\theta}\thinspace d\theta= \frac{1}{2}e^{2\theta}\sin{3\theta}-\frac{3}{4}e^{2\theta}\cos{3\theta}$$ $$= \frac{2}{13}e^{2\theta}\sin{3\theta}-\frac{3}{13}e^{2\theta}\cos{3\theta}+C$$
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