Question

Evaluate the integral.

\(\int_0^{\sqrt 3 /2} {\frac{{dr}}{{\sqrt {1 - {r^2}} }}} \).

Short answer

The solution of given integral\(\int_0^{\sqrt 3 /2} {\frac{{dr}}{{\sqrt {1 - {r^2}} }}} \) is \(\frac{\pi }{3}\).

Step-by-step solution

Definition of integration or Antiderivative

The function\({\rm{F}}\)is called as antiderivative of\({\rm{f}}\)on an interval\({\rm{I}}\)if\({\rm{F'(x) = f(x)}}\)

i.e.,\(\int {{\rm{f(x)dx = F(x)}}} \).

\(\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = {\sin ^{ - 1}}x + C\)

Applying formula of integration

Using evaluation theorem writes given integral as-

\(\begin{aligned}{c}\int_0^{\sqrt 3 /2} {\frac{{dr}}{{\sqrt {1 - {r^2}} }}} = \left( {{{\sin }^{ - 1}}x} \right)_0^{\sqrt 3 /2}\\ = {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}(0)\\ = \frac{\pi }{3} - 0\\ = \frac{\pi }{3}\end{aligned}\)

Therefore, the solution of given integral\(\int_0^{\sqrt 3 /2} {\frac{{dr}}{{\sqrt {1 - {r^2}} }}} \) is \(\frac{\pi }{3}\).