Answer
$$\frac{\theta }{2}\sin \left( {2\theta + 1} \right) + \frac{1}{4}\cos \left( {2\theta + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\theta \cos \left( {2\theta + 1} \right)} d\theta \cr
& {\text{Using the method of integration by parts}} \cr
& \,\,\,\,\,\,\,\,\,{\text{Let }}\,\,\,\,\,u = \theta ,\,\,\,\,\,du = d\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \cos \left( {2\theta + 1} \right),\,\,\,\,\,v = \frac{1}{2}\sin \left( {2\theta + 1} \right) \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \cr
& \to \,\,\,\int {\theta \cos \left( {2\theta + 1} \right)} d\theta = \theta \left[ {\frac{1}{2}\sin \left( {2\theta + 1} \right)} \right] - \int {\frac{1}{2}\sin \left( {2\theta + 1} \right)d\theta } \cr
& = \frac{\theta }{2}\sin \left( {2\theta + 1} \right) - \frac{1}{2}\int {\sin \left( {2\theta + 1} \right)d\theta } \cr
& {\text{Integrate and simplify}} \cr
& = \frac{\theta }{2}\sin \left( {2\theta + 1} \right) - \frac{1}{2}\left( { - \frac{1}{2}\cos \left( {2\theta + 1} \right)} \right) + C \cr
& = \frac{\theta }{2}\sin \left( {2\theta + 1} \right) + \frac{1}{4}\cos \left( {2\theta + 1} \right) + C \cr} $$
