Calculus: Repeated Integration by Parts

Formula: $\int u \, dv = u v - \int v \, du$. Let $u = x^2 \implies du = 2x \, dx$ Let $dv = \cos(2x) \, dx \implies v = \frac{1}{2}\sin(2x)$ $\int x^2 \cos(2x) \, dx = \frac{1}{2} x^2 \sin(2x) - \int \frac{1}{2}\sin(2x) \cdot 2x \, dx = \frac{1}{2} x^2 \sin(2x) - \int x \sin(2x) \, dx$

For $\int x \sin(2x) \, dx$: Let $u_1 = x \implies du_1 = dx$ Let $dv_1 = \sin(2x) \, dx \implies v_1 = -\frac{1}{2}\cos(2x)$ $\int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) - \int \left(-\frac{1}{2}\cos(2x)\right) dx = -\frac{1}{2} x \cos(2x) + \frac{1}{4}\sin(2x)$

$\int x^2 \cos(2x) \, dx = \frac{1}{2} x^2 \sin(2x) - \left( -\frac{1}{2} x \cos(2x) + \frac{1}{4}\sin(2x) \right) + C$ $= \frac{1}{2} x^2 \sin(2x) + \frac{1}{2} x \cos(2x) - \frac{1}{4}\sin(2x) + C$