Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Practice Exercises - Page 307: 26


$-6 \sqrt 3$

Work Step by Step

Consider $f(x)=\int_{-(\pi/3)}^{(\pi/3)} (\sec^2 x-8 \cos x) dx$ Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ As we know that $\int_{-k}^k f(x) dx=2 \int_0^{k} f(x) dx$ This implies that $2[\tan x-8 \sin x]_{0}^{\pi/3}=2[(\tan (\dfrac{\pi}{3})-\tan (0))-8 (\sin (\dfrac{\pi}{3})-\sin (0))=2\sqrt 3-8\sqrt 3=\sqrt 3 (2-8)$ Hence, $f(x)=\int_{-(\pi/3)}^{(\pi/3)} (\sec^2 x-8 \cos x) dx=-6 \sqrt 3$
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.