Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 522: 71

Answer

$$\int e^{a x} \sin b x d x=\frac{e^{a x}(a \sin b x-b \cos b x)}{a^{2}+b^{2}}+C$$

Work Step by Step

Let $$I= \int e^{a x} \sin b x d x $$ Use integration by parts , let \begin{aligned} u&= \sin b x \ \ \ \ \ \ &dv&= e^{a x} dx\\ du&= b\cos b x dx \ \ \ \ \ \ &v&=\frac{1}{a} e^{a x} \end{aligned} then \begin{aligned} I &= uv-\int vdu \\ &= \frac{1}{a} e^{a x} \sin bx-\frac{b}{a}\int e^{a x}\cos bxdx \\ & = \frac{1}{a} e^{a x} \sin bx-\frac{b}{a}J\ \ \ \ \ \ \ (1)\\ \end{aligned} Where $$J=\int e^{a x}\cos bxdx$$ Use integration by parts , let \begin{aligned} u&= \cos b x \ \ \ \ \ \ &dv&= e^{a x} dx\\ du&= -b\sin b x dx \ \ \ \ \ \ &v&=\frac{1}{a} e^{a x} \end{aligned} Then \begin{aligned} J&= uv-\int vdu\\ &= \frac{1}{a} e^{a x} \cos bx+ \frac{b}{a}\int e^{ax} \sin bxdx\\ &= \frac{1}{a} e^{a x} \cos bx+ \frac{b}{a}I\ \ \ \ \ \ \ \ \ \ \ (2) \end{aligned} It follows that from (1) and (2) \begin{aligned} I & = \frac{1}{a} e^{a x} \sin bx-\frac{b}{a}J\\ & = \frac{1}{a} e^{a x} \sin bx-\frac{b}{a}\left[\frac{1}{a} e^{a x} \cos bx+ \frac{b}{a}I\right]\\ &= \frac{1}{a} e^{a x} \sin bx-\frac{b}{a^2}e^{a x} \cos bx-\frac{b^2}{a^2}I\\ \left(1+\frac{b^2}{a^2}\right)I&= e^{a x}[\frac{1}{a}\sin bx -\frac{b}{a^2}\cos bx]+C\\ \left( \frac{a^2+b^2}{a^2}\right)I&= e^{a x}[\frac{1}{a}\sin bx -\frac{b}{a^2}\cos bx]+C\\ \left( a^2+b^2 \right)I&= e^{a x}[a\sin bx -b\cos bx]+C\\ \end{aligned} Hence $$\int e^{a x} \sin b x d x=\frac{e^{a x}(a \sin b x-b \cos b x)}{a^{2}+b^{2}}+C$$
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