Thomas' Calculus 13th Edition

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

Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 28

Answer

$$2-\sqrt{2}$$

Work Step by Step

We integrate as follows: \begin{align*} \int_{0}^{\pi / 6} \sqrt{1+\sin x} d x&=\int_{0}^{\pi / 6} \frac{\sqrt{1+\sin x}}{1} d x\\ &=\int_{0}^{\pi / 6} \frac{\sqrt{1-\sin ^{2} x}}{\sqrt{1-\sin x}} d x\\ &=\int_{0}^{\pi / 6} \frac{\sqrt{\cos ^{2} x}}{\sqrt{1-\sin x}} d x\\ &=\int_{0}^{\pi / 6} \frac{\cos x}{\sqrt{1-\sin x}} d x \\ &= -2(1-\sin x)^{1 / 2}\bigg|_{0}^{\pi / 6}\\ &=-2 \sqrt{1-\sin \left(\frac{\pi}{6}\right)}+2 \sqrt{1-\sin 0}\\ &=-2 \sqrt{\frac{1}{2}}+2 \sqrt{1}\\ &=2-\sqrt{2} \end{align*}
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