Let $$\left[\cdot\right]$$ be the greatest integer function. If $$(\alpha = \int_{0}^{64} \left( x^{1/3} - [x^{1/3}] \right)\, dx $$, then $$\frac{1}{\pi} \int_{0}^{\alpha\pi } \left( \frac{\sin^{2}\theta } {\sin^{6}\theta + \cos^{6}\theta} \right) d\theta$$ is equal to ____ .

$$\alpha=\int_0^{64}\left(x^{1/3}-\lfloor x^{1/3}\rfloor\right)dx$$

$$Put(x=t^3\Rightarrow dx=3t^2dt,;t:0\to4):$$
$$\alpha=\int_0^4(t-\lfloor t\rfloor)3t^2dt$$
$$=\sum_{n=0}^3\int_n^{n+1}3t^2(t-n)dt=36$$

Now:
$$\frac{1}{\pi}\int_0^{\alpha\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}d\theta$$
$$=\frac{1}{\pi}\int_0^{36\pi}$$...

Period $$(=\pi),$$ so:
=36$$.\frac{1}{\pi}\int_0^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}d\theta$$

Using symmetry:
$$\int_0^{\pi}\frac{\sin^2\theta}{\sin^6+\cos^6}d\theta$$
$$=\int_0^{\pi}\frac{\cos^2\theta}{\sin^6+\cos^6}d\theta$$
$$\Rightarrow\text{each}=\frac{\pi}{2}$$

Hence value =36