Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 49



Work Step by Step

To evaluate the integral $$\int_{1/2}^1\frac{\cos(x^{-2})}{x^3}dx$$ we will use substitution $x^{-2}=t$ which gives us $-2x^{-3}dx=dt\Rightarrow\frac{dx}{x^3}=-\frac{dt}{2}$ and the integration bounds would be: for $x=1/2$ we have $t=4$ and for $x=1$ we have $t=1$. Putting this into the integral we get: $$\int_{1/2}^1\frac{\cos(x^{-2})}{x^3}dx=\int_4^1\cos t\left(-\frac{dt}{2}\right)= -\frac{1}{2}\int_4^1\cos t dt=\frac{1}{2}\int_1^4\cos t dt= \left.\frac{1}{2}\sin t\right|_1^4=\frac{1}{2}(\sin4-\sin1)$$
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