Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx \cr
& {\text{Use the substitution method}}{\text{. }} \cr
& u = 1 + 3{\sin ^2}x,\,\,\,\,du = 6\sin x\cos xdx,\,\,\,\,\,dx = \frac{{du}}{{6\sin x\cos x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx = \int {\frac{{3\sin x\cos x}}{{\sqrt u }}} \left( {\frac{{du}}{{6\sin x\cos x}}} \right) \cr
& = \frac{1}{2}\int {\frac{{du}}{{\sqrt u }}} \cr
& = \frac{1}{2}\int {{u^{ - 1/2}}du} \cr
& {\text{Integrate by the power rule}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \sqrt u + C \cr
& {\text{Write in terms of }}x;{\text{ replace }}u = 1 + 3{\sin ^2}x \cr
& = \sqrt {1 + 3{{\sin }^2}x} + C \cr
& \cr
& \int_0^{\pi /2} {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx = \left( {\sqrt {1 + 3{{\sin }^2}x} } \right)_0^{\pi /2} \cr
& {\text{Evaluate}} \cr
& = \sqrt {1 + 3{{\sin }^2}\left( {\frac{\pi }{2}} \right)} - \sqrt {1 + 3{{\sin }^2}\left( 0 \right)} \cr
& = \sqrt {1 + 3} - \sqrt {1 + 0} \cr
& = 2 - 1 \cr
& = 1 \cr} $$
