Multiplication of Taylor series - expanding $2x\sin(x)$

Your approach is correct. WolframAlpha returns the same series for the following query: Series[2 x Sin[x],{x,0,8}]

In order to estimate the integral of $2x\sin x$, you use the series for $\sin$ as you wrote above, multiplied by $2$ and $x$ as you did.

Thus

$$2x\sin x =2\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!}= 2x^2 - \frac{2x^4}{3!} + \frac{2x^6}{5!} - \frac{2x^8}{7!} \cdots $$

We then integrate the summand or term-by-term to get

$$ 2\int x\sin x\, dx= 2\sum_{n=0}^{\infty} \int\frac{(-1)^n x^{2n+2}}{(2n+1)!}\,dx= 2\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!}\int x^{2n+2}\,dx= 2\sum_{n=0}^{\infty}\frac{(-1)^n x^{3+2 n}}{(3+2 n) (1+2 n)!} $$

This sum can be used to estimate the integral of $2x\sin x$ to a desired accuracy.

To check your result,

$$2\int x\sin x=-2 (x \cos(x)-\sin(x))$$