Answer
$$\frac{2}{5}\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 1/5}^{1/5} {\frac{{6dx}}{{\sqrt {4 - 25{x^2}} }}} \cr
& = \int_{ - 1/5}^{1/5} {\frac{{6dx}}{{\sqrt {{2^2} - {{\left( {5x} \right)}^2}} }}} \cr
& {\text{Use the substitution method}}{\text{:}} \cr
& u = 5x,{\text{ so that }}du = 5dx \cr
& {\text{The new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1/5,{\text{ then }}u = 5\left( {1/5} \right) = 1 \cr
& \,\,\,\,\,\,{\text{If }}x = - 1/5,{\text{ then }}u = 5\left( { - 1/5} \right) = - 1 \cr
& {\text{then}} \cr
& \int_{ - 1/5}^{1/5} {\frac{{6dx}}{{\sqrt {{2^2} - {{\left( {5x} \right)}^2}} }}} = \int_{ - 1}^1 {\frac{{6\left( {du/5} \right)}}{{\sqrt {{2^2} - {u^2}} }}} \cr
& = \frac{6}{5}\int_{ - 1}^1 {\frac{{du}}{{\sqrt {{2^2} - {u^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \frac{6}{5}\left( {{{\sin }^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_{ - 1}^1 \cr
& = \frac{6}{5}\left( {{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) - {{\sin }^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)} \right) \cr
& = \frac{6}{5}\left( {\frac{\pi }{6} + \frac{\pi }{6}} \right) \cr
& = \frac{2}{5}\pi \cr} $$
