Answer
$$\frac{1}{9}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt 3 /2} {\frac{4}{{9 + 4{x^2}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& \int_0^{\sqrt 3 /2} {\frac{4}{{9 + 4{x^2}}}} dx = 2\int_0^{\sqrt 3 /2} {\frac{2}{{{{\left( 3 \right)}^2} + {{\left( {2x} \right)}^2}}}} dx \cr
& {\text{Let }}u = 2x,\,\,\,du = 2dx \cr
& {\text{Apply the substitution}} \cr
& 2\int_0^{\sqrt 3 /2} {\frac{2}{{{{\left( 3 \right)}^2} + {{\left( {2x} \right)}^2}}}} dx = 2\int_{2\left( 0 \right)}^{2\left( {\sqrt 3 /2} \right)} {\frac{{du}}{{{{\left( 3 \right)}^2} + {u^2}}}} \cr
& = 2\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{u}{3}} \right)} \right)_0^{\sqrt 3 } \cr
& {\text{Evaluate}} \cr
& {\text{ = }}\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{3}} \right) - \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{0}{3}} \right) \cr
& = \frac{1}{9}\pi \cr} $$
