Answer
$$\frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{3\sqrt 2 /4} {\frac{{ds}}{{\sqrt {9 - 4{s^2}} }}} \cr
& = \int_0^{3\sqrt 2 /4} {\frac{{ds}}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {2s} \right)}^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 2s,{\text{ so that }}du = 2ds \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}s = 3\sqrt 2 /4,{\text{ then }}u = 2\left( {3\sqrt 2 /4} \right) = 3\sqrt 2 /2 \cr
& \,\,\,\,\,\,{\text{If }}s = 0,{\text{ then }}u = 2\left( 0 \right) = 0 \cr
& {\text{then}} \cr
& \int_0^{3\sqrt 2 /4} {\frac{{ds}}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {2s} \right)}^2}} }}} = \int_0^{3\sqrt 2 /2} {\frac{{du/2}}{{\sqrt {{{\left( 3 \right)}^2} - {u^2}} }}} \cr
& = \frac{1}{2}\int_0^{3\sqrt 2 /2} {\frac{{du}}{{\sqrt {{{\left( 3 \right)}^2} - {u^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 3 \cr
& = \frac{1}{2}\left( {{{\sin }^{ - 1}}\left( {\frac{u}{3}} \right)} \right)_0^{3\sqrt 2 /2} + C \cr
& = \frac{1}{2}\left( {{{\sin }^{ - 1}}\left( {\frac{{3\sqrt 2 }}{{3\left( 2 \right)}}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{3}} \right)} \right) \cr
& = \frac{1}{2}\left( {\frac{\pi }{4} - 0} \right) \cr
& = \frac{\pi }{8} \cr} $$
