Calculate Value of $\alpha$

We need to calculate $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$. The term $x^{1/3} - [x^{1/3}]$ represents the fractional part of $x^{1/3}$. We evaluate the integral by splitting the interval $[0, 64]$ based on the integer values of $x^{1/3}$.

The values of $x$ where $[x^{1/3}]$ changes are $x=1^3=1$, $x=2^3=8$, $x=3^3=27$, and $x=4^3=64$.

• For $0 \le x < 1$, $[x^{1/3}] = 0$. The integral is $\int_{0}^{1} (x^{1/3} - 0) dx$.
• For $1 \le x < 8$, $[x^{1/3}] = 1$. The integral is $\int_{1}^{8} (x^{1/3} - 1) dx$.
• For $8 \le x < 27$, $[x^{1/3}] = 2$. The integral is $\int_{8}^{27} (x^{1/3} - 2) dx$.
• For $27 \le x \le 64$, $[x^{1/3}] = 3$. The integral is $\int_{27}^{64} (x^{1/3} - 3) dx$.

The antiderivative of $x^{1/3}$ is $\frac{3}{4} x^{4/3}$.

• $\int_{0}^{1} x^{1/3} dx = [\frac{3}{4} x^{4/3}]_{0}^{1} = \frac{3}{4}$.
• $\int_{1}^{8} (x^{1/3} - 1) dx = [\frac{3}{4} x^{4/3} - x]_{1}^{8} = (\frac{3}{4} (16) - 8) - (\frac{3}{4} - 1) = (12 - 8) - (-\frac{1}{4}) = 4 + \frac{1}{4} = \frac{17}{4}$.
• $\int_{8}^{27} (x^{1/3} - 2) dx = [\frac{3}{4} x^{4/3} - 2x]_{8}^{27} = (\frac{3}{4} (81) - 54) - (\frac{3}{4} (16) - 16) = (\frac{243}{4} - \frac{216}{4}) - (12 - 16) = \frac{27}{4} - (-4) = \frac{27}{4} + \frac{16}{4} = \frac{43}{4}$.
• $\int_{27}^{64} (x^{1/3} - 3) dx = [\frac{3}{4} x^{4/3} - 3x]_{27}^{64} = (\frac{3}{4} (256) - 192) - (\frac{3}{4} (81) - 81) = (192 - 192) - (\frac{243}{4} - \frac{324}{4}) = 0 - (-\frac{81}{4}) = \frac{81}{4}$.

Summing these values:

$\alpha = \frac{3}{4} + \frac{17}{4} + \frac{43}{4} + \frac{81}{4} = \frac{3 + 17 + 43 + 81}{4} = \frac{144}{4} = 36$.

Evaluate the Second Integral

We need to evaluate $\frac{1}{\pi} \int_{0}^{\alpha \pi} \left(\frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}\right) d\theta$. Substituting $\alpha=36$, the expression becomes $\frac{1}{\pi} \int_{0}^{36 \pi} \left(\frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}\right) d\theta$.

Let $g(\theta) = \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}$. The function $g(\theta)$ has a period of $\pi$, since $g(\theta + \pi) = g(\theta)$.

The integral over $36\pi$ can be written as:

$\int_{0}^{36 \pi} g(\theta) d\theta = 36 \times \int_{0}^{\pi} g(\theta) d\theta$.

Also, $g(\theta)$ is symmetric about $\theta = \pi/2$ in the interval $[0, \pi]$, meaning $g(\pi - \theta) = g(\theta)$. Therefore:

$\int_{0}^{\pi} g(\theta) d\theta = 2 \times \int_{0}^{\pi/2} g(\theta) d\theta$.

Let's evaluate $I_{p/2} = \int_{0}^{\pi/2} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta$. Divide the numerator and denominator by $\cos^6 \theta$:

$I_{p/2} = \int_{0}^{\pi/2} \frac{\tan^2 \theta \sec^2 \theta}{\tan^6 \theta + 1} d\theta$.

Use the substitution $t = \tan \theta$. Then $dt = \sec^2 \theta d\theta$. The limits change from $[0, \pi/2]$ to $[0, \infty)$.

$I_{p/2} = \int_{0}^{\infty} \frac{t^2}{t^6 + 1} dt$.

Now, let $u = t^3$. Then $du = 3t^2 dt$, which means $t^2 dt = \frac{1}{3} du$. The limits remain $[0, \infty)$.

$I_{p/2} = \int_{0}^{\infty} \frac{1}{u^2 + 1} \left(\frac{1}{3} du\right) = \frac{1}{3} \int_{0}^{\infty} \frac{1}{u^2 + 1} du$.

The integral of $\frac{1}{u^2 + 1}$ is $\arctan(u)$.

$I_{p/2} = \frac{1}{3} [\arctan(u)]_{0}^{\infty} = \frac{1}{3} (\lim_{u \to \infty} \arctan(u) - \arctan(0)) = \frac{1}{3} (\frac{\pi}{2} - 0) = \frac{\pi}{6}$.

Now, we find the integral over $[0, \pi]$:

$\int_{0}^{\pi} g(\theta) d\theta = 2 \times I_{p/2} = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

The integral over $[0, 36\pi]$ is:

$\int_{0}^{36 \pi} g(\theta) d\theta = 36 \times \frac{\pi}{3} = 12 \pi$.

Finally, the expression we need to evaluate is:

$\frac{1}{\pi} \int_{0}^{36 \pi} g(\theta) d\theta = \frac{1}{\pi} (12 \pi) = 12$.

Final Result

The calculated value of the expression is 12.