I tried to compute $\int_0^1x^2\sin^2xd x$ and mathematica disagrees with me I tried to compute $\int_0^1x^2\sin^2xd x$ and got $\frac 16-\frac 38\sin2-\frac 14\cos2$. However, Wolfram Mathematica gives the result $\frac 16-\frac 18\sin2-\frac 14\cos2$, namely there is something wrong with the coefficient of $\sin$ in my answer. And here is how I computed it(If you don't have the patience to read it through, please help me do the integral and see if we get the same answer):
$$\int_0^1x^2\sin^2xd x$$
$$=\int^1_0x^2\frac {1-\cos 2x}{2}dx$$
$$=\int^1_0\frac {x^2}2 dx+\int^1_0\frac{-x^2\cos 2x}2dx$$
$$=\frac 12*\frac 13[x^3]^1_{x=1_00}+\int^1_0\frac{-x^2\cos 2x}{2}\frac{d(\sin2x)}{2\cos 2x}$$
$$=\frac 16+\int^1_0\frac{-x^2}{4}d(\sin 2x)$$
$$=\frac 16+((\frac {-x^2}{4}\sin 2x)^1_0-\int^1_0\sin 2xd(\frac {-x^2}{4}))$$
$$=\frac 16+(-\frac 14\sin 2+\frac 14\int^1_0\sin 2x*2xdx)$$
$$\frac 16-\frac 14\sin 2+\frac 12\int^1_0x\sin 2xdx$$
But $$\int^1_0x\sin2xdx=\int^1_0x\sin2x\frac{d\cos2x}{-2\sin 2x}=-\frac 12\int^1_0xd(\cos 2x)=-\frac 12((x\cos 2x)^1_0+\int^1_0-\cos 2x\frac {d(2x)}{2})=-\frac 12(\cos 2+\frac 12(\sin 2x)^1_0)=-\frac12(\cos 2+\frac 12\sin 2)$$, So $$\int^1_0x^2\sin^2xdx=\frac 16-\frac 14\sin 2+\frac 12(-\frac 12)(\cos 2+\frac 12\sin 2)=\frac 16-\frac 14 \sin 2-14\cos 2-\frac 18\sin2=\frac 16-\frac 38\sin 2-\frac 14\cos 2$$
 A: $$\int x^2\cos(2x)\,dx = \text{Re}\int x^2 e^{2ix}\,dx =\frac{1}{4}\text{Re}\int x^2 e^{ix}\,dx=\frac{1}{4}\text{Re}\left((2i+2x-ix^2)e^{ix}\right)$$
hence:
$$\int_{0}^{1} x^2\cos(2x)\,dx = \frac{1}{4}\left(2\cos 2+\sin 2\right).$$
A: $$\int^1_0x\sin2xdx=\int^1_0x\sin2x\frac{d\cos2x}{-2\sin 2x}=-\frac 12\int^1_0xd(\cos 2x)=-\frac 12((x\cos 2x)^1_0\color{#C00}{+}\int^1_0-\cos 2x\frac {d(2x)}{2})=-\frac 12(\cos 2\color{#C00}{+}\frac 12(\sin 2x)^1_0)=-\frac12(\cos 2\color{#C00}{+}\frac 12\sin 2)$$
If I'm not mistaken, this should be a minus sign, which gives Mathematica's (correct) result.

EDIT:
$$\int^1_0x\sin2xdx = \left[-\frac12x\cos2x\right]^1_0 - \int^1_0-\frac12\cos2xdx = -\frac12\cos2 - \left[-\frac14\sin2x\right]^1_0 = -\frac12\cos2 + \frac14\sin2$$
A: Using the identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$ along with tabular integration, you should get $$\begin{align}\int_0^1 x^2\sin^2(x) = \frac{1}{2}\left(\int^1_0x^2dx-\int_0^1x^2\cos(2x) \right) \\ =  \frac{1}{2}\left(\frac{x^3}{3}\Biggl\vert_0^1-\int_0^1x^2\cos(2x) \right) \\ = \frac{1}{2}\left(\frac{1}{3}-\left[\frac{x^2}{2}\sin(2x)+\frac{x}{2}\cos(2x)-\frac{1}{4}\sin(2x)\right]_0^1 \right) \\ =  \frac{1}{2}\left(\frac{1}{3}-\left[\frac{1}{2}\sin(2)+\frac{1}{2}\cos(2)-\frac{1}{4}\sin(2)\right] \right) \\ = \frac{1}{2}\left(\frac{1}{3}-\left[\frac{1}{4}\sin(2)+\frac{1}{2}\cos(2)\right] \right) \\ = \frac{1}{6}-\frac{1}{8}\sin(2)-\frac{1}{4}\cos(2)  \end{align}$$ which is exactly what Wolfram gives for a result.
