Answer
$$\sqrt {17} - 4$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\frac{{\cos \theta \sin \theta }}{{\sqrt {{{\cos }^2}\theta + 16} }}} d\theta \cr
& {\text{set }}u = \cos \theta ,\,\,\,\,du = - \sin \theta d\theta \cr
& {\text{switch the limits of integration using }}u = \cos \theta \cr
& \theta = 0 \to 1 \cr
& \theta = \pi /2 \to 0 \cr
& {\text{use the substitution}} \cr
& \int_0^{\pi /2} {\frac{{\cos \theta \sin \theta }}{{\sqrt {{{\cos }^2}\theta + 16} }}} d\theta = \int_1^0 {\frac{{ - udu}}{{\sqrt {{u^2} + 16} }}} \cr
& \cr
& {\text{use the substitution }}v = {u^2} + 16,\,\,\,\,\,\,\,\,dv = 2udu,\,\,\,\,\,\frac{1}{2}dv = udu \cr
& {\text{switch the limits of integration using }}v = {u^2} + 16 \cr
& u = 0 \to v = 16 \cr
& u = 1 \to v = 17 \cr
& {\text{use the substitution}} \cr
& \int_1^0 {\frac{{ - udu}}{{\sqrt {{u^2} + 16} }}} = - \int_{17}^{16} {\frac{{\left( {1/2} \right)dv}}{{\sqrt v }}} \cr
& = - \frac{1}{2}\int_{17}^{16} {{v^{ - 1/2}}} dv \cr
& {\text{integrate}} \cr
& = - \frac{1}{2}\left( {\frac{{{v^{1/2}}}}{{1/2}}} \right)_{17}^{16} + C \cr
& = - \left( {\sqrt v } \right)_{17}^{16} + C \cr
& {\text{evaluate the limits}} \cr
& = - \sqrt {16} + \sqrt {17} \cr
& = \sqrt {17} - 4 \cr} $$
