Answer
$$\frac{{3\pi }}{{16}}$$
Work Step by Step
$$\eqalign{
& \,\,\,\,\int_0^{\pi /2} {{{\sin }^4}\theta } d\theta \cr
& = \int_0^{\pi /2} {{{\left( {{{\sin }^2}\theta } \right)}^2}} d\theta \cr
& {\text{use the half - angle formula }}{\sin ^2}x = \frac{1}{2} - \frac{{\cos x\theta }}{2} \cr
& \int_0^{\pi /2} {{{\left( {{{\sin }^2}\theta } \right)}^2}} d\theta = \int_0^{\pi /2} {{{\left( {\frac{1}{2} - \frac{{\cos 2\theta }}{2}} \right)}^2}} d\theta \cr
& = \frac{1}{4}\int_0^{\pi /2} {{{\left( {1 - \cos 2\theta } \right)}^2}} d\theta \cr
& {\text{expanding the binomial}} \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 - 2\cos 2\theta + {{\cos }^2}2\theta } \right)} d\theta \cr
& {\text{use the half - angle formula }}{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 - 2\cos 2\theta + \frac{{1 + \cos 4\theta }}{2}} \right)} d\theta \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 - 2\cos 2\theta + \frac{1}{2} + \frac{{\cos 4\theta }}{2}} \right)} d\theta \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {\frac{3}{2} - 2\cos 2\theta + \frac{{\cos 4\theta }}{2}} \right)} d\theta \cr
& {\text{integrate}} \cr
& = \frac{1}{4}\left[ {\frac{3}{2}\theta - \sin 2\theta + \frac{1}{8}\sin 4\theta } \right]_0^{\pi /2} \cr
& {\text{evaluating the limits}} \cr
& = \frac{1}{4}\left[ {\frac{3}{2}\left( {\frac{\pi }{2}} \right) - \sin 2\left( {\frac{\pi }{2}} \right) + \frac{1}{8}\sin 4\left( {\frac{\pi }{2}} \right)} \right] - \frac{1}{4}\left[ {\frac{3}{2}\left( 0 \right) - \sin 2\left( 0 \right) + \frac{1}{8}\sin 4\left( 0 \right)} \right] \cr
& {\text{simplifying}} \cr
& = \frac{1}{4}\left[ {\frac{3}{4}\pi - 0 + 0} \right] - \frac{1}{4}\left[ {0 - 0 + 0} \right] \cr
& = \frac{{3\pi }}{{16}} \cr} $$
