## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 30

#### Answer

$$\int^{3\pi/4}_{\pi/2}\sqrt{1-\sin 2x}dx=1$$

#### Work Step by Step

$$A=\int^{3\pi/4}_{\pi/2}\sqrt{1-\sin 2x}dx$$ Multiply both numerator and denominator with $\sqrt{1+\sin 2x}$ $$A=\int^{3\pi/4}_{\pi/2}\sqrt{1-\sin 2x}\times\frac{\sqrt{1+\sin 2x}}{\sqrt{1+\sin 2x}}dx$$ $$A=\int^{3\pi/4}_{\pi/2}\frac{\sqrt{1-\sin^22x}}{\sqrt{1+\sin 2x}}dx$$ $$A=\int^{3\pi/4}_{\pi/2}\frac{\sqrt{\cos^22x}}{\sqrt{1+\sin 2x}}dx$$ $$A=\int^{3\pi/4}_{\pi/2}\frac{-\cos 2x}{\sqrt{1+\sin 2x}}dx$$ (because for $x\in[\pi/2,3\pi/4]$, we have $2x\in[\pi,3\pi/2]$, on whose interval $\cos 2x\lt0$) We set $u=\sqrt{1+\sin 2x}$, which means $$du=\frac{2\cos 2x}{2\sqrt{1+\sin 2x}}dx=\frac{\cos2x}{\sqrt{1+\sin2x}}dx$$ - For $x=3\pi/4$, $u=\sqrt{1+\sin(3\pi/2)}=\sqrt{1-1}=0$ - For $x=\pi/2$, $u=\sqrt{1+\sin\pi}=1$ $$A=\int^0_1-du=\int^1_0du$$ $$A=u\Big]^1_0$$ $$A=1$$

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