Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 41


$$\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)dx=0$$

Work Step by Step

Denote the function under the integral $f(x)=x^3+x^4\tan x.$ Now we have $$f(-x)=(-x)^3+(-x)^4\tan(-x)=-x^3+x^4(-\tan x)=-(x^3+x^4\tan x)=-f(x)$$ i.e. the function $f$ is odd. Knowing that the integral of an odd function with the symmetric integration bounds (one bound is $-a$ and the other is $a$) is equal to $0$ we have that: $$\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)dx=0$$
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