Animated Solution for Mathematics - Definite Integration: ${\lim }_{x\to \frac{\pi }{2}}\left(\frac{1}{(x-\frac{\pi }{2}{)}^2}{\int }_{x^3}^{(\frac{\pi }{2}{)}^3}\cos (t^{\frac{1}{3}})dt\right)$ is equal to

Welcome, future engineers and mathematicians. Today, we are going to dissect a problem that, at first glance, looks like a monster. We have a limit involving an integral with a variable limit, and it is sitting right there in the numerator.

In the world of JEE Advanced, intimidation is just a test of your fundamentals. Let us break this down, step by step, and find the beauty hidden within the algebra.

Before we touch any theorem, we must perform the sacred ritual of limits: direct substitution. We are evaluating:

If we plug in $x=\frac{\pi }{2}$, the lower limit of our integral becomes $(\frac{\pi }{2}{)}^3$, which is exactly equal to the upper limit. As we know, the integral of any function from $a$ to $a$ is zero.

The denominator also vanishes to zero. We have a $0/0$ indeterminate form. This is our green light to proceed with L'Hôpital's Rule.

Now, how do we differentiate an integral? This is where the Newton-Leibniz Rule becomes our best friend. It states:

Applying this to our numerator, the upper limit is a constant, so its derivative is zero. The lower limit is $x^3$, so we substitute $t=x^3$ into our function $\cos (t^{1/3})$, giving us $\cos ((x^3{)}^{1/3})=\cos (x)$.

Then, we multiply by the derivative of the lower limit, which is $3x^2$. Don't forget the negative sign from the formula! Our numerator becomes $-3x^2\cos (x)$.

We differentiate the denominator, $(x-\frac{\pi }{2}{)}^2$, using the chain rule to get $2(x-\frac{\pi }{2})$. Putting it all together, we have:

If we check the form again, we still have $0/0$. We could differentiate again, but that would involve the product rule and get messy. Let's be smarter.

We need to relate $\cos (x)$ to $(x-\frac{\pi }{2})$. Recall the identity $\cos (x)=\sin (\frac{\pi }{2}-x)$.

Since $\sin (-\theta )=-\sin (\theta )$, we can rewrite this as $\cos (x)=-\sin (x-\frac{\pi }{2})$. Substituting this back into our limit, the negative signs cancel out beautifully!

We are left with:

Now, look at the structure. We have $\frac{\sin (x-\frac{\pi }{2})}{x-\frac{\pi }{2}}$. As $x\to \frac{\pi }{2}$, this is the classic standard limit ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }=1$.

The rest of the expression, $\frac{3x^2}{2}$, is perfectly well-behaved. Substituting $x=\frac{\pi }{2}$, we get:

And there it is—the elegance of calculus in action. The final answer is $\frac{3{\pi }^2}{8}$. Keep practicing, stay curious, and remember that every complex problem is just a series of simple steps waiting to be connected.