Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 68


$\frac{\sqrt 3}{2}-1+\frac{\pi}{6}$

Work Step by Step

We have $$\int_{-1/2}^0\frac{(x+1)dx}{\sqrt{1-x^2}}=\int_{-1/2}^0\frac{x dx}{\sqrt{1-x^2}}+\int_{-1/2}^0\frac{dx}{\sqrt{1-x^2}}\\ =-\frac{1}{2}\int_{-1/2}^0\frac{-2x dx}{\sqrt{1-x^2}}+\int_{-1/2}^0\frac{dx}{\sqrt{1-x^2}}\\ =-\frac{1}{2}\frac{ \sqrt{1-x^2}}{1/2}|_{-1/2}^0+\sin^{-1}x|_{-1/2}^0\\ =-1+\frac{\sqrt 3}{2}+0+\frac{\pi}{6}\\=\frac{\sqrt 3}{2}-1+\frac{\pi}{6} .$$
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