University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 14

Answer

$$\int\frac{1}{x^2}\cos^2\frac{1}{x}dx=-\frac{1}{2x}-\frac{1}{4}\sin\frac{2}{x}+C$$

Work Step by Step

$$A=\int\frac{1}{x^2}\cos^2\frac{1}{x}dx$$ We set $u=-1/x$. Then $$du=-\Big(-\frac{1}{x^2}\Big)dx=\frac{1}{x^2}dx$$ Also, $$\frac{1}{x}=-u$$ Therefore, $$A=\int\cos^2(-u)du=\int(\cos(-u))^2du$$ We have, according to trigonometric identities, $\cos(-u)=\cos u$. So $(\cos(-u))^2=\cos^2u$ We also have an identity stating that $\cos^2u=\frac{1+\cos2u}{2}$ Therefore, $$A=\frac{1}{2}\int(1+\cos2u)du$$ $$A=\frac{1}{2}\Big(u+\frac{1}{2}\sin2u\Big)+C$$ $$A=\frac{1}{2}\Big(-\frac{1}{x}+\frac{1}{2}\sin\Big(-\frac{2}{x}\Big)\Big)+C$$ $$A=\frac{1}{2}\Big(-\frac{1}{x}-\frac{1}{2}\sin\frac{2}{x}\Big)+C$$ (since $\sin(-x)=-\sin x$)$$A=-\frac{1}{2x}-\frac{1}{4}\sin\frac{2}{x}+C$$
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