## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 24

#### Answer

$$\int e^{-2x}\sin2xdx=-\frac{e^{-2x}}{4}(\sin2x+\cos2x)+C$$

#### Work Step by Step

$$A=\int e^{-2x}\sin2xdx$$ Set $u=\sin2x$ and $dv=e^{-2x}dx$ Then we have $du=2\cos 2xdx$ and $v=-\frac{1}{2}e^{-2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=-\frac{1}{2}e^{-2x}\sin2x-\int-\frac{1}{2}e^{-2x}\times2\cos2xdx$$ $$A=-\frac{1}{2}e^{-2x}\sin2x+\int e^{-2x}\cos2xdx$$ Set $u=\cos2x$ and $dv=e^{-2x}dx$ Then we have $du=-2\sin 2xdx$ and $v=-\frac{1}{2}e^{-2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=-\frac{1}{2}e^{-2x}\sin2x+\Big(-\frac{1}{2}\Big)e^{-2x}\cos2x-\int\Big(-\frac{1}{2}\Big)e^{-2x}\times(-2)\sin2xdx$$ $$A=-\frac{1}{2}e^{-2x}\sin2x-\frac{1}{2}e^{-2x}\cos2x-\int e^{-2x}\sin2xdx$$ We notice that above, $\int e^{-2x}\sin2xdx$ is exactly the given integral to solve, meaning it equals $A$. Therefore, $$A=-\frac{1}{2}e^{-2x}\sin2x-\frac{1}{2}e^{-2x}\cos2x-A$$ $$2A=-\frac{1}{2}e^{-2x}\sin2x-\frac{1}{2}e^{-2x}\cos2x+C$$ $$A=-\frac{1}{4}e^{-2x}\sin2x-\frac{1}{4}e^{-2x}\cos2x+C$$ $$A=-\frac{e^{-2x}}{4}(\sin2x+\cos2x)+C$$

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