Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 427: 99

Answer

$\frac{\pi^2}{4}$

Work Step by Step

Using substitution $u = \pi - x\,\,\,du=-dx$, we have $\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\, dx = \int_{0}^{\pi}\frac{(\pi-u)\sin u}{1+\cos^2u}\, du$ There is no change to $\sin x $ and $\cos x$ since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=(-\cos x)^2=\cos^2x$, at the same time $x$ and $u$ are dummy variables which means they are interchangeable, therefore $\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\, dx =\frac{\pi}{2} \int_{0}^{\pi}\frac{\sin x}{1+\cos^2x}\, dx$ Now, by substitution $t=\cos x\,\,\,dt=-\sin x\, dx$, thus $\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\, dx =\frac{\pi}{2} \int_{0}^{\pi}\frac{\sin x}{1+\cos^2x}\, dx=\frac{\pi}{2}\int_{-1}^{1}\frac{1}{1+t^2}\,\,\, dt=\frac{\pi}{2}\biggl[\arctan t\biggr]_{-1}^{1}=\frac{\pi^2}{4}$ Remark : $\int\frac{1}{1+x^2}\, dx=\arctan x +C$
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.