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: 18

Answer

$-\frac{4}{3}\cos^3(\frac{1}{2}x)+C$

Work Step by Step

$\int\sin x\cos(\frac{1}{2}x)dx$ Let $u=\frac{1}{2}x$. Then $x=2u$. Also, $du=\frac{1}{2}dx$, and $dx=2du$. $=\int\sin(2u)\cos u*2\ du$ $=2\int 2\sin u\cos u*\cos u\ du$ $=4\int\sin u\cos^2 u\ du$ Let $v=\cos u$. Then $dv=-\sin u\ du$, and $\sin u\ du=-dv$. $=4\int v^2*-1 dv$ $=-4*\frac{v^3}{3}+C$ $=-\frac{4}{3}\cos^3 u+C$ $=\boxed{-\frac{4}{3}\cos^3(\frac{1}{2}x)+C}$
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