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: 1

Answer

$\frac{1}{3}$$sin^{3}$x-$\frac{1}{5}$$sin^{5}$x+C

Work Step by Step

$\int$$sin^{2}$x$cos^{3}$xdx =$\int$$sin^{2}$x$cos^{2}$xcosxdx, split the odd power =$\int$$sin^{2}$x(1-$sin^{2}$x)cosxdx, make $cos^{2}$=(1-$sin^{2}$x) Now use u-sub: u=sinx; du=cosxdx =$\int$$u^{2}$(1-$u^{2}$)du =$\int$($u^{2}$-$u^{4}$)du =$\frac{1}{3}$$u^{3}$x-$\frac{1}{5}$$u^{5}$x+C =$\frac{1}{3}$$sin^{3}$x-$\frac{1}{5}$$sin^{5}$x+C
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