Animated Solution for Mathematics - Definite Integration: If $f(x)={\int }_0^{x}t(\sin x-\sin t)dt$ then :

Mastering the Leibniz Rule

Welcome, future engineer. Today, we are going to tackle a problem that separates the casual student from the true master of calculus. We are looking at the function $f(x)={\int }_0^{x}t(\sin x-\sin t)dt$.

At first glance, it looks like a standard integral. But look closer. Do you see the trap?

The variable $x$ is not just sitting comfortably in the upper limit of the integral; it is also lurking inside the integrand, disguised as $\sin x$. This is a classic JEE Advanced setup, designed to test if you know when to use the standard Fundamental Theorem of Calculus and when to reach for the more powerful Leibniz Integral Rule.

The Leibniz Rule – Your Secret Weapon

When you see a function defined as an integral where the variable of differentiation appears both in the limits and inside the integrand, do not panic. Do not try to force a standard integration.

Instead, invoke the Leibniz Integral Rule. Think of this rule as a scalpel; it allows us to differentiate under the integral sign with surgical precision. The formula is:

It looks intimidating, but it is just a three-part process. We differentiate the inside, we account for the upper limit, and we account for the lower limit. Let us apply this to our function $f(x)$.

The First Derivative – The Aha! Moment

Let us find $f^{\prime }(x)$. First, we take the partial derivative of the integrand $t(\sin x-\sin t)$ with respect to $x$. Since we are treating $t$ as a constant during this partial differentiation, $t\sin t$ vanishes, and we are left with $t\cos x$.

Now, for the boundary terms. The upper limit is $x$, so we evaluate the integrand at $t=x$, yielding $x(\sin x-\sin x)$, which is $0$. The lower limit is $0$, and evaluating there also yields $0$.

The boundary terms vanish completely! We are left with:

Since $\cos x$ does not depend on $t$, we pull it out of the integral. The integral of $t$ is simply $\frac{t^2}{2}$. Evaluating from $0$ to $x$, we get:

We have successfully crossed the first hurdle. Take a breath. You have just tamed a complex integral.

The Chain of Derivatives – The Grind

Now, we need the second and third derivatives. This is where patience is your greatest virtue. To find $f^{\prime \prime }(x)$, we differentiate $\frac{x^2}{2}\cos x$ using the product rule:

See how the terms evolve? Now, for the third derivative, $f^{\prime \prime \prime }(x)$. We apply the product rule again, this time to both terms:

Subtracting these, we get:

The Beauty of Cancellation

Let us simplify this expression. Distributing the negative sign, we have:

Now, look at the last term: $-\frac{x^2}{2}\cos x$. Does it look familiar? Look back at our result for $f^{\prime }(x)$. It is exactly $-\frac{x^2}{2}\cos x=-f^{\prime }(x)$.

This is the moment of clarity. We substitute this back into our equation:

Rearranging to match our final form:

And there it is. The math aligns perfectly. You didn't just solve a problem; you navigated a logical path, respected the rules of calculus, and found the hidden symmetry. Keep this mindset—the ability to see the structure within the chaos—and you will conquer any problem the JEE throws at you.