Question

Evaluate the definite integral: $∫_0^1 \frac{dx}{√1-x^2}$

Answer 1

Let I=$∫_0^1 \frac{dx}{√1-x^2}$

$∫\frac{dx}{√1-x^2}=sin^{-1} x=f(x)$

By second fundamental theorem of calculus,we obtain

$I=F(1)-F(0)$

=sin-1(1)-sin-1(0)

$=\frac{π}{2}-0$

$=\frac{π}{2}$