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Today's calculation of Integral 635

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August 14, 2010
calculusintegrationfunctiontrigonometryderivativecalculus computations

Problem Statement

Suppose that a function f(x)f(x) defined in −1<x<1-1<x<1 satisfies the following properties (i) , (ii), (iii).
(i) f′(x)f'(x) is continuous.
(ii) When −1<x<0, f′(x)<0, f′(0)=0-1<x<0,\ f'(x)<0,\ f'(0)=0, when 0<x<1, f′(x)>00<x<1,\ f'(x)>0.
(iii) f(0)=−1f(0)=-1
Let F(x)=∫0x1+{f′(t)}2dt (−1<x<1)F(x)=\int_0^x \sqrt{1+\{f'(t)\}^2}dt\ (-1<x<1). If F(sin⁡θ)=cθ (c:constant)F(\sin \theta)=c\theta\ (c :\text{constant}) holds for −π2<θ<π2-\frac{\pi}{2}<\theta <\frac{\pi}{2}, then find f(x)f(x).
1975 Waseda University entrance exam/Science and Technology
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