Prove that $\left|\int_{-2\pi}^{2\pi}x^{2}\sin^{8}\left(e^{x}\right)dx\right|\le\frac{16\pi^{3}}{3}.$

Question

Prove that $\left|\int_{-2\pi}^{2\pi}x^{2}\sin^{8}\left(e^{x}\right)dx\right|\le\frac{16\pi^{3}}{3}.$

I know that $\left|\int_{-2\pi}^{2\pi}x^{2}\sin^{8}\left(e^{x}\right)dx\right| \le \int_{-2\pi}^{2\pi}|x^{2}\sin^{8} \left(e^{x}\right)dx| \le U(|f|,P) \le U(f,P)$ for any partition $P$ which would be less then or equal to $(2\pi-(-2\pi))(2\pi)^2\sin^8(e^{2\pi})\le \frac{16\pi^3}{3}$

Since $\sin^8(e^{2\pi})$ is greater then $\frac{1}{3}$.

Answer 1

Hint. One has $$ \left|\sin^8 (e^x)\right|\le 1,\quad x \in \mathbb{R}. $$