Answer
\[ = - \frac{1}{2}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\frac{\pi }{2}} {x\cos 2xdx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
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v = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = dx \hfill \\
dv = \cos 2x\,\,\, \to \,\,\,\,\,\,\,v = \frac{1}{2}\sin 2x \hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
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{\text{replacing the values }}{\text{in the equation}} \hfill \\
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{\text{ = }}\,\frac{x}{2}\sin 2x - \frac{1}{2}\int_{}^{} {\sin 2xdx} \hfill \\
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or \hfill \\
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{\text{ = }}\,\frac{x}{2}\sin 2x - \frac{1}{4}\int_{}^{} {\sin 2x\left( 2 \right)dx} \hfill \\
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integrate \hfill \\
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= \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + C \hfill \\
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{\text{Therefore}}{\text{,}} \hfill \\
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\int_0^{\frac{\pi }{2}} {x\cos 2xdx} = \,\,\left[ { = \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x} \right]_0^{\frac{\pi }{2}} \hfill \\
\hfill \\
evaluate\,\,the\,\,limits \hfill \\
\hfill \\
= \,\,\left[ {\frac{\pi }{4}\sin \,\left( \pi \right) + \frac{1}{4}\cos \,\left( \pi \right)} \right] - \,\,\left[ {0 + \frac{1}{2}} \right] \hfill \\
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{\text{Simplify}} \hfill \\
\hfill \\
= - \frac{1}{4} - \frac{1}{4} = - \frac{1}{2} \hfill \\
\end{gathered} \]
