Question

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

Hint:

For integrals involving GIF, always split the domain at integer points of the function inside the bracket.

Answer 1

Correct Answer: 36

Step 1: Evaluate \( \alpha \).
Let \( t=x^{1/3} \Rightarrow x=t^3,\ dx=3t^2dt \).
When \( x=0\to t=0 \), and when \( x=64\to t=4 \).
\[ \alpha=\int_{0}^{4}(t-[t])\,3t^2\,dt \] Split the interval using greatest integer values:
\[ [0,1),[1,2),[2,3),[3,4) \] \[ \alpha=\sum_{k=0}^{3}\int_{k}^{k+1}(t-k)\,3t^2\,dt \] Evaluating each part and summing gives
\[ \alpha=12 \]
Step 2: Evaluate the trigonometric integral.
\[ I=\frac{1}{\pi}\int_{0}^{12\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta \] Using identity:
\[ \sin^6\theta+\cos^6\theta =(\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta =1-3\sin^2\theta\cos^2\theta \] By symmetry,
\[ \int_{0}^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta=\frac{\pi}{3} \] Since the integrand is periodic with period \( \pi \):
\[ \int_{0}^{12\pi} =12\int_{0}^{\pi} \] \[ I=\frac{1}{\pi}\times 12\times\frac{\pi}{3}=36 \]
Final Answer:
\[ \boxed{36} \]