f(t)dt Using Leibnitz Rule, √1−(f′(x))2=f(x) ∴f′(x)=±√1−f2(x)⇒∫‌
df(x)
√1−f2(x)
=±∫dx sin−1(f(x))=±x+c ∴f(x)=sin(±x+C) ‌∵f(0)=0⇒C=0 ∴f(x)=sin‌x‌ or ‌−sin‌x ‌ But ‌f‌ is non-negative on ‌[0,π∕2] ∴f(x)=‌ Sin ‌x ‌∵f(x)<x‌ for all ‌x>0 ∴f(‌