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

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: $$-\frac{1}{4}\cos{(2)}\cdot x^2+\frac{1}{8}\sin{(2x^2)}+C$$

• Is this the original integral or did you do some work to get here? – Michael Burr Apr 3 '16 at 12:04
• It is the original integral. I must say that I've come across more mistakes in this workbook I'm using, so there is a chance of a mistake, eather in the beginning integral or solution. – A6SE Apr 3 '16 at 12:07
• This is doable with $(1-x^2)$ and I doubt it is with $(1-x^3)$. – Yves Daoust Apr 3 '16 at 12:09
• Hmm, wow, it could be, I'll try to solve now. Thank you. – A6SE Apr 3 '16 at 12:11
• Even with $1-x^2$ the answer is different. – mathemather Apr 3 '16 at 12:16

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