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Today's calculation of integral 104

Source: Kyoto Prefectural University of Medicine entrance exam1989

June 5, 2006

calculusintegrationtrigonometryfunctioncalculus computations

Problem Statement

For

0<x<1,

let

f(x)=\int_0^x \frac{dt}{\sqrt{1-t^2}}\ dt

(1) Find

\frac{d}{dx} f(\sqrt{1-x^2})

(2) Find

f\left(\frac{1}{\sqrt{2}}\right)

(3) Prove that

f(x)+f(\sqrt{1-x^2})=\frac{\pi}{2}