Computing an integral via integration by parts

Question

Good afternoon,

I am learning for Monday's test and I stuck on this example:

$$ \int (2x+3) \, \cos{2x} \, \mathrm{d}x $$

and I am asked to determine this integral. I am trying to do it by integration by parts, but it still leads me nowhere… Could you guys help me out with this?

Thank you

Answer 1

Let $u=2x+3$, $dv=\cos(2x)\,dx$ so that $du=2\,dx$, $v={1\over 2}\sin(2x)$. Then \begin{align*} \int (2x+3)\cos(2x)\,dx&=(2x+3)\cdot{1\over 2}\sin(2x)-\int {1\over 2}\sin(2x)\cdot 2\,dx\\ &=(2x+3)\cdot{1\over 2}\sin(2x)-\int \sin(2x)\,dx\\ &=(2x+3)\cdot{1\over 2}\sin(2x)+{1\over 2}\cos(2x)+C. \end{align*}