We start with taking the constant factor 5 out of the integral, which leaves us with
#5 int x cos(2x)\ dx#.
If we look at the integrand now, then we have a factor #x#, whose derivative is #1#, and a factor #cos(2x)#, which is super-easy to integrate with an anti-derivative of #sin(2x) / 2#. Therefore, this calls for integration by parts, which is described by the general rule
#int u' v = u v - int u v'#.
In our particular case, we choose #u = cos(2x)# and #v = x#, and thus get
#5 int x cos(2x)\ dx = \frac{5}{2} [ x sin(2x) - int sin(2x)\ dx ]#.
The last remaining integral is again an easy one,
#int sin(2x)\ dx = - cos(2x)/2#,
and so we arrive at the final result
#\frac{5}{2} [ x sin(2x) - int sin(2x)\ dx ] =#
#= \frac{5}{2} [ x sin(2x) + cos(2x)/2 ] + C =#
#= \frac{5}{4} [ 2 x sin(2x) + cos(2x) ] + C#.
