Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 19

Answer

\begin{aligned} \int 16 \sin ^{2} x \cos ^{2} x d x=2 x-4 \sin x \cos ^{3} x+2 \sin x \cos x+C \end{aligned}

Work Step by Step

Given $$ \int 16 \sin ^{2} x \cos ^{2} x d x $$ So, we have \begin{aligned} I&= \int 16 \sin ^{2} x \cos ^{2} x d x\\ &\text{since} \ \ \sin ^{2} x=\frac{1-\cos 2 x}{2} \ \ \cos ^{2}x=\frac{1+\cos 2 x}{2} ,\text{ we get}\\ &=16 \int\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1+\cos 2 x}{2}\right) d x\\ &=4 \int\left(1-\cos ^{2} 2 x\right) d x\\ &=4 \int\left( \sin ^{2} 2 x\right) d x\\ &\text{since} \ \ \sin ^{2} 2 x=\frac{1-\cos 4 x}{2} \ \ ,\text{ we get}\\ &= 4 \int\left(\frac{1-\cos 4 x}{2}\right) d x\\ &=2 \int d x-\frac{1}{2} \int 4\cos 4 x d x\\ &=2 x-\frac{1}{2} \sin 4 x+C\\ &=2 x-\frac{1}{2} \sin 4 x+C\\ &=2 x-\sin 2 x \cos 2 x+C\\ &=2 x-2 \sin x \cos x\left(2 \cos ^{2} x-1\right)+C\\ &=2 x-4 \sin x \cos ^{3} x+2 \sin x \cos x+C \end{aligned}
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