Animated Solution for Mathematics - Definite Integration: Let $[\cdot ]$ be the greatest integer function. If $\alpha ={\int }_0^{64}(x^{1/3}-[x^{1/3}])dx$, 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 .........

Welcome, JEE warrior. Today, we are not just solving a problem; we are embarking on a journey through the landscape of calculus. This problem is a masterpiece—a perfect blend of the discrete nature of the Greatest Integer Function and the continuous, rhythmic beauty of trigonometry.

Take a deep breath. Let us dismantle this beast, piece by piece.

The Staircase of the GIF

We begin with $\alpha ={\int }_0^{64}(x^{1/3}-[x^{1/3}])dx$. At first glance, this looks like a standard integral, but the presence of the Greatest Integer Function (GIF) tells us that the function is not smooth. It is a staircase.

The function $[x^{1/3}]$ remains constant as long as $x^{1/3}$ is between two consecutive integers. Imagine walking along the $x$-axis. As you pass $x=1$, $x=8$, $x=27$, and $x=64$, the value of $[x^{1/3}]$ jumps.

We must respect these jumps. We split the integral into four distinct regions: $[0,1)$, $[1,8)$, $[8,27)$, and $[27,64)$. In each region, $[x^{1/3}]$ is a constant integer.

For instance, in the interval $[8,27)$, $x^{1/3}$ ranges from $2$ to $3$, so $[x^{1/3}]=2$. The area contribution is simply $2\times (27-8)=38$. By summing these areas, we find the total area under the GIF to be $156$.

Subtracting this from the integral of $x^{1/3}$:

We arrive at $\alpha =192-156=36$. We have conquered the first peak!

The Trigonometric Dance

Now, we face the second part:

The limit $36\pi$ is intimidating, but in the world of JEE, a large limit is often a gift. It signals periodicity. We checked the function $f(\theta )=\frac{{\sin }^2\theta }{{\sin }^6\theta +{\cos }^6\theta }$ and discovered its period is $\pi$.

This allows us to collapse the integral:

The $\pi$ in the denominator cancels out, leaving us with $36{\int }_0^{\pi }f(\theta )d\theta$. The problem is shrinking before our eyes.

The Algebraic Alchemy

To solve ${\int }_0^{\pi }\frac{{\sin }^2\theta }{{\sin }^6\theta +{\cos }^6\theta }d\theta$, we need to bridge the gap between trigonometry and algebra. We divide the numerator and denominator by ${\cos }^6\theta$.

The integrand transforms into:

This is the moment of clarity. By substituting $u=\tan \theta$, we convert the trigonometric integral into an algebraic one. The limits change from $[0,\pi ]$ to $[0,\infty )$ (with a factor of $2$ due to symmetry about $\pi /2$).

We are left with:

The Grand Finale

We factorize the denominator $u^6+1$ as $(u^2+1)(u^4-u^2+1)$. The $(u^2+1)$ terms cancel beautifully, leaving us with:

To solve this, we divide by $u^2$ to get:

By substituting $t=u-\frac{1}{u}$, we get $dt=(1+\frac{1}{u^2})du$. The integral transforms into the standard form:

Multiplying by our constant $36$, we reach the final answer: 36. It was a long road, but look at the elegance of the result. Every step was necessary, every substitution was a key turning in a lock. You have mastered the logic.