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 347: 66

Answer

\begin{aligned} \int_{0}^{\pi / 2} \cos^2 x d x&=\frac{\pi}{4}\\ \int_{0}^{\pi / 2} \sin ^2x d x&= \frac{\pi}{4} \end{aligned}

Work Step by Step

Since we have $$\int_{0}^{\pi / 2} f(\cos x) d x=\int_{0}^{\pi / 2} f(\sin x) d x$$ Then $$\int_{0}^{\pi / 2} \cos^2 x d x=\int_{0}^{\pi / 2} \sin ^2x d x$$ Use the rule $$\cos^2x +\sin^2x=1 $$ We get \begin{aligned} \int_{0}^{\pi / 2} \cos^2 x d x+ \int_{0}^{\pi / 2} \sin ^2x d x&=\int_{0}^{\pi / 2} (\cos^2x+\sin ^2x) d x\\ &= \int_{0}^{\pi / 2} dx\\ &= x\bigg|_{0}^{\pi / 2}\\ &=\frac{\pi}{2} \end{aligned} Hence \begin{aligned} \int_{0}^{\pi / 2} \cos^2 x d x&=\frac{\pi}{4}\\ \int_{0}^{\pi / 2} \sin ^2x d x&= \frac{\pi}{4} \end{aligned}
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