## Calculus: Early Transcendentals (2nd Edition)

$= \frac{{x{{\sin }^2}x}}{2} - \frac{x}{4} - \frac{{\sin 2x}}{8} + C$
$\begin{gathered} \int_{}^{} {x\sin x\cos x\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = x\sin x\,\,\,\,\,\, \to \,\,\,\,du = x\cos + \sin x \hfill \\ dv = \cos x\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,v = \sin x \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \,\left( {x\sin x} \right)\,\left( {\sin x} \right) - \int_{}^{} {\sin x\,\left( {x\cos x + \sin x} \right)} \,dx \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ x{\sin ^2}x - \int_{}^{} {x\sin x\cos x\,dx - \int_{}^{} {{{\sin }^2}x\,dx} } \hfill \\ \hfill \\ = 2\int_{}^{} {x\sin x\cos x\,dx = x{{\sin }^2}x - \int_{}^{} {{{\sin }^2}x\,dx} } \hfill \\ \hfill \\ use\,\,the\,\,identity\,\,{\sin ^2}x\, = \frac{{1 - \cos 2x}}{2} \hfill \\ \hfill \\ = 2\int_{}^{} {x\sin xdx = x{{\sin }^2}x - \int_{}^{} {\,\left( {\frac{{1 - \cos 2x}}{2}} \right)} } \,dx \hfill \\ or \hfill \\ = 2\int_{}^{} {x\sin xdx = x{{\sin }^2}x - \int_{}^{} \, } \,dx + \int_{}^{} {\,\left( {\frac{{\cos 2x}}{2}} \right)} dx \hfill \\ \hfill \\ inetgrate \hfill \\ \hfill \\ 2\int_{}^{} {x\sin xdx = x{{\sin }^2}x - \frac{x}{2} + \frac{{\sin 2x}}{4} + C} \hfill \\ \hfill \\ solve\,for\,\,\int_{}^{} {x\sin xdx} \hfill \\ \hfill \\ \,\,\int_{}^{} {x\sin x\cos xdx} = \frac{{x{{\sin }^2}x}}{2} - \frac{x}{4} - \frac{{\sin 2x}}{8} + C \hfill \\ \end{gathered}$