Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 380: 22

$\frac{\pi}{4}$

$\int_{0}^{\sqrt 2}\frac{1}{\sqrt (4-x^{2})}dx$ The denominator $\sqrt (4-x^{2})$ looks like $arcsin$. The formula to solve to $arcsin$ is: $\int\frac{du}{a^{2}-u^{2}}=arcsin(\frac{u}{a})+C$ $u=x$ $du=dx$ $a=2$ $[arcsin(\frac{x}{2}]_{0}^{\sqrt 2}$ $[arcsin(\frac{\sqrt 2}{2})-arcsin(\frac{0}{2})]$ $[\frac{\pi}{4}-0]$ $\frac{\pi}{4}$