## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 18

#### Answer

$$y' = -cos(x)\sqrt{1+sin^2x}$$

#### Work Step by Step

$$y=\int\limits_{sinx}^1 {\sqrt{1 + t^2}}dt$$ Switch the bounds, making the integral negative $$y=-\int\limits^{sinx}_1 {\sqrt{1 + t^2}}dt$$ Substitute the upper bound in for t and multiply by the derivative. $$y' = -\sqrt{1 + sin^2(x)} \times (sinx)'$$ Simplify. $$y' = -\sqrt{1 + sin^2(x)} \times (cosx)$$ $$y' = -cos(x)\sqrt{1+sin^2x}$$

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