University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 93


$$\int^{\pi/2}_0\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx=1$$

Work Step by Step

$$A=\int^{\pi/2}_0\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx$$ We set $u=\sqrt{1+3\sin^2x}$, which means $$du=\frac{(1+3\sin^2x)'}{2\sqrt{1+3\sin^2x}}dx=\frac{6\sin x\cos x}{2\sqrt{1+3\sin^2x}}dx$$ $$du=\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx$$ For $x=\pi/2$: $$u=\sqrt{1+3\sin^2(\pi/2)}=\sqrt{1+3\times1^2}=2$$ For $x=0$: $$u=\sqrt{1+3\sin^20}=\sqrt{1+3\times0^2}=1$$ Therefore, $$A=\int^2_1du$$ $$A=u\Big]^2_1$$ $$A=2-1=1$$
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