Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises: 15

Answer

$\ln |\sin x|-\frac{1}{2}\sin^2 x+C$

Work Step by Step

$\int\cot x\cos^2 x\ dx$ $=\int\frac{\cos x}{\sin x}*\cos^2 x\ dx$ $=\int\frac{\cos^3 x}{\sin x}\ dx$ $=\int\frac{\cos x*\cos^2 x}{\sin x}\ dx$ $=\int\frac{\cos x(1-\sin^2 x)}{\sin x}\ dx$ Let $u=\sin x$. Then $du=\cos x\ dx$. $=\int\frac{1-u^2}{u}du$ $=\int(\frac{1}{u}-\frac{u^2}{u})du$ $=\int(u^{-1}-u)du$ $=\ln |u|-\frac{u^2}{2}+C$ $=\boxed{\ln |\sin x|-\frac{1}{2}\sin^2 x+C}$
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