Trying to evaluate this indefinite integral:
$ \int (x^2 + 1)\cos2xdx$
So far I have the following: $u=x^2 + 1 \Rightarrow du = 2xdx$ and $dv=\cos2x \Rightarrow v = \frac {\sin2x}{2}$. So the integral is equal to:
$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\int {\frac{\sin2x}{2}}2xdx$
Next, I make another substitution for the integral on the right hand side; let $ u = x \Rightarrow du = dx$ and let $dv = \sin2x \Rightarrow v = \frac {-\cos2x}{2}$. Now I have the following:
$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left (-\frac {x\cos2x}{2} - \int -\frac {cos2x}{2}dx\right)$
Which after integrating becomes:
$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left(-\frac {x\cos2x}{4} + \frac {\sin2x}{4}\right)$
But when solving with the integrator on my calculator, I get a different answer (it looks like I am getting closer, but still off). What am I doing wrong here??