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