Tips on how to integrate $\int x\sin{(1+x^2)}\sin{(1-x^3)}\ dx$

Question

So, I'm completely stuck at integrating this integral: $$\int x\sin{(1+x^2)}\sin{(1-x^3)}\ dx$$ The solution should supposedly be: $-\frac{1}{4}x^2\cos{(2x^2)}+\frac{1}{8}\sin{(2x^2)}+C$.

Using trigonometric identity: $$\sin(\alpha)\sin(\beta)=\frac{1}{2}\bigg[\cos{(\alpha-\beta)}-\cos{(\alpha+\beta)}\bigg]$$ I get: $$\int x\sin{(1+x^2)}\sin{(1-x^3)}\ dx=\frac{1}{2}\int x\cos{(x^3+x^2)}\ dx\ -\ \frac{1}{2}\int x\cos{(x^3-x^2-2)}\ dx$$ But, I have no idea what to do with those integrals.
I've also tried using: $$\sin{(\alpha\pm\beta)}=\sin{(\alpha)}\cos{(\beta)}\pm\cos{(\alpha)}\sin{(\beta)}$$ But, it led me nowhere. Any tips on how to solve this? Thank you for your time.

Edit: So, there are mistakes in both the integral and solution. The original integral should be (thanks Yves Daoust): $$\int x\sin{(1+x^2)}\sin{(1-x^2)}\ dx$$ The solution should be: $$-\frac{1}{4}\cos{(2)}\cdot x^2+\frac{1}{8}\sin{(2x^2)}+C$$

Answer 1

The derivative of the supposed answer is $$ -\frac{1}{2}x\cos(2x^2)+x^3\sin(2x^2)+\frac{1}{2}x\cos(2x^2)=x^3\sin(2x^2) $$ Substituting $x=1$ into this expression gives $\sin(2)$, while the original expression is $0$ when $x=1$. Therefore, there appears to be an error with the supposed solution.

Answer 2

With $u=x^2$,

$$4\int x \sin(1+x^2)\sin(1-x^2)dx=2\int \sin(1+u)\sin(1-u)du=\int(\cos(2u)-\cos(2))du.$$

You can finish from here.