University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 29

Answer

$$\int\sin(\ln x)dx=\frac{x}{2}(\sin\ln x-\cos\ln x)+C$$

Work Step by Step

$$A=\int\sin(\ln x)dx$$ Let $a=\ln x$, which means $x=e^a$ Then $da=\frac{1}{x}dx=\frac{1}{e^a}dx$. So $dx=e^ada$ $$A=\int\sin a\times e^ada=\int e^a\sin ada$$ Let $u=\sin a$ and $dv=e^ada$ We will have $du=\cos ada$ and $v=e^a$ Apply the formula $\int udv=uv-\int vdu$: $$A=e^a\sin a-\int e^a\cos ada$$ We carry out integration by parts one more time. Let $u=\cos a$ and $dv=e^ada$ We will have $du=-\sin ada$ and $v=e^a$ Apply the formula $\int udv=uv-\int vdu$: $$A=e^a\sin a-\Big(e^a\cos a-\int -e^a\sin ada\Big)$$ $$A=e^a\sin a-\Big(e^a\cos a+\int e^a\sin ada\Big)$$ We notice that $\int e^a\sin ada$ is exactly the given integral to solve, which means it equals $A$. $$A=e^a\sin a-\Big(e^a\cos a+A\Big)$$ $$A=e^a\sin a-e^a\cos a-A$$ $$2A=e^a\sin a-e^a\cos a+C=e^a(\sin a-\cos a)+C$$ $$A=\frac{e^a}{2}(\sin a-\cos a)+C$$ Replace $a$ back to $\ln x$: $$A=\frac{e^{\ln x}}{2}(\sin\ln x-\cos\ln x)+C$$ $$A=\frac{x}{2}(\sin\ln x-\cos\ln x)+C$$
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