$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx 4$ Evaluate
$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx </4$
I tried to use the fact that the $\sin^3x<\sin x$. It does not help. This qesion is proposed by Jalil Hajimir to RMM.

There is more than likely a much simpler solution.
Use the fact that, over the considered range the cube of the sine is smaller than the cube of its argument. Developing the later and using a couple of integration by parts plus triple angle formulae, you end for
$$I=\int \left(x+\sin3x+\cos3x\right)^3\,dx$$
$$I=\frac{x^4}{4}+\frac{3 x^2}{2}+x^2 \sin (3 x)-x^2 \cos (3 x)+\frac{2}{3} x \sin (3
   x)+\frac{5}{18} \sin (3 x)+\frac{1}{12} \sin (6 x)-\frac{1}{18} \sin (9
   x)+\frac{2}{3} x \cos (3 x)-\frac{1}{2} x \cos (6 x)-\frac{5}{18} \cos (3
   x)-\frac{1}{18} \cos (9 x)$$ and, using the bounds
$$J=\int_{\frac \pi 6}^{\frac \pi 2} \left(x+\sin3x+\cos3x\right)^3\,dx=\frac{ 5 \pi ^4+18 \pi ^2 -90 \pi-216} {324}\approx 0.512207$$ which is even smaller than $\frac \pi 6$.
