Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 8


$$\frac{2x+\sin \left(2x\right)-4\cos ^3\left(x\right)\sin \left(x\right)}{16}+C$$

Work Step by Step

Given $$\int \sin ^{2} x \cos ^{2} x d x$$ Since \begin{align*} \int \sin ^{2} x \cos ^{2} x d x&=\int (1-\cos ^{2} x ) \cos ^{2} x d x\\ &=\int (\cos ^{2} x -\cos ^{4} x ) d x\\ \end{align*} Use $$ \int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$$ Then \begin{align*} \int \sin ^{2} x \cos ^{2} x d x&=\int (1-\cos ^{2} x ) \cos ^{2} x d x\\ &=\int (\cos ^{2} x -\cos ^{4} x ) d x\\ &=\frac{\cos x \sin x}{2}+\frac{1}{2} \int d x-\frac{\cos ^{3} x \sin x}{4}-\frac{3}{4} \int \cos ^{2} x d x\\ &=\frac{\cos x \sin x}{2}+\frac{1}{2} x-\frac{\cos ^{3} x \sin x}{4}-\frac{3}{4}\left(\frac{\cos x \sin x}{2}+\frac{1}{2} x\right)+C \\ &=\frac{2x+\sin \left(2x\right)-4\cos ^3\left(x\right)\sin \left(x\right)}{16}+C \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.