## Calculus: Early Transcendentals (2nd Edition)

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