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 - 8.1 Exercises - Page 513: 74


The required formula is $$\int \frac{1}{z^2}dz=-\frac{1}{z}+c.$$

Work Step by Step

The integral $$\int x \sec(x^2+1)\tan(x^2+1)dx=\int x\frac{\sin(x^2+1)}{\cos^2(x^2+1)}dx$$ with the substitution $t=x^2+1$ where is $dt=2xdx$ can be transformed into integral $$\frac{1}{2}\int \frac{\sin t}{\cos^2t}dt.$$ Finally with the substitution $z=\cos t$ we get integral $$-\frac{1}{2}\int\frac{1}{z^2}dz,$$ so the required formula is $$\int \frac{1}{z^2}dz=-\frac{1}{z}+c.$$
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