## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{3}$$
\eqalign{ & \int_0^{\sqrt {\pi /2} } {x{{\sin }^3}\left( {{x^2}} \right)} dx \cr & {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr & {\text{express the limits in terms of }}u \cr & x = \sqrt {\pi /2} {\text{ implies }}u = {\left( {\sqrt {\pi /2} } \right)^2} = \pi /2 \cr & x = 0{\text{ implies }}u = {\left( 0 \right)^2} = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & = \frac{1}{2}\int_0^{\pi /2} {{{\sin }^3}u} du \cr & {\text{split off }}{\sin ^3}u \cr & = \frac{1}{2}\int_0^{\pi /2} {{{\sin }^2}u} \sin udu \cr & {\text{pythagorean identity}} \cr & = \frac{1}{2}\int_0^{\pi /2} {\left( {1 - {{\cos }^2}u} \right)} \sin udu \cr & = \frac{1}{2}\int_0^{\pi /2} {\left( {\sin u - {{\cos }^2}u\sin u} \right)} du \cr & {\text{evaluate}} \cr & = \frac{1}{2}\left[ { - \cos u + \frac{{{{\cos }^3}u}}{3}} \right]_0^{\pi /2} \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{2}\left[ { - \cos \left( {\frac{\pi }{2}} \right) + \frac{{{{\cos }^3}\left( {\frac{\pi }{2}} \right)}}{3}} \right] - \frac{1}{2}\left[ { - \cos \left( 0 \right) + \frac{{{{\cos }^3}\left( 0 \right)}}{3}} \right] \cr & {\text{simplify}} \cr & = \frac{1}{2}\left[ 0 \right] - \frac{1}{2}\left[ { - 1 + \frac{1}{3}} \right] \cr & = \frac{1}{3} \cr}