Integrate $x^2\sin(2x)$ using $u$-substitution

Question

I want to integrate $$\int x^2(sin(2x))dx$$ using $u$-substitution.

Seeing that $x^2$ and $2x$ made me think that $u$ should equal $x^2$ and working through it I immediately got stuck at $$ \int u\left(sin(du)\right)$$ cause I think we all know that just ain't right. So instead I used $u=2x$ $$\frac{1}{2}\int \sqrt{\frac{u}{2}}sin(u)du$$ but it seems I'm just over complicating the problem, so which $u$-sub is correct?

Answer 1

$\int x^{2} \sin (2x)dx =-\frac 1 2 \cos (2x)x^{2} +\int x\cos (2x)dx$. Another integration by parts gives $\int x^{2} \sin (2x)dx =-\frac 1 2 \cos (2x)x^{2}+\frac 1 2 x\sin (2x) -\frac 1 2\int \sin (2x) dx$. So the answer is $-\frac 1 2 \cos (2x)x^{2}+\frac 1 2 x\sin (2x) dx+\frac 1 4 \cos (2x)+C$ where $C$ is a constant.

Answer 2

As $\sin 2x = \Im(e^{2ix})$

$$I = \int x^2\sin(2x)dx = \Im\int x^2e^{2ix}dx = \Im\left(x^2\frac{e^{2ix}}{2i}-2x\frac{e^{2ix}}{4i^2}+2\frac{e^{2ix}}{8i^3}\right)+C$$

$$I = -x^2\frac{\cos 2x}{2}+\frac{x}{2}\sin 2x+\frac{1}{4}\cos2x+C=\frac{(1-2x^2)\cos 2x+2x\sin 2x}{4}+C$$

I've used Bernoulli's formula

$\int uv dx = uv_1 - u'v_2 + u''v_3 \cdots$