Animated Solution for Mathematics - Definite Integration: The value of ${\int }_1^{e^{37}}\frac{\pi \sin (\pi \ln x)}{x}dx$ is $\dots \dots \dots$

Welcome, warrior of JEE. Today, we face an integral that, at first glance, looks like a formidable beast. You see

and your heart might skip a beat. But in the world of advanced calculus, what looks like a monster is often just a simple problem wearing a mask. Let us unmask it together.

The secret to conquering this integral lies in recognizing a pattern. Look at the integrand: we have a sine function with a complex argument, $\pi \ln x$, and sitting right there in the denominator is $x$.

Your mathematical intuition should immediately trigger a response: the derivative of $\ln x$ is $\frac{1}{x}$. This is the classic function-derivative pair that screams for substitution.

Let us define $t=\pi \ln x$. When we differentiate this with respect to $x$, we get:

Look at that! The $\frac{\pi }{x}dx$ is exactly what we have in our integral. The monster is already starting to shrink.

This is the moment where many students stumble, but you will not. When we change our variable from $x$ to $t$, we must also change our boundaries.

Our original lower limit is $x=1$. Plugging this into our substitution $t=\pi \ln x$, we get $t=\pi \ln (1)=0$. Our new lower limit is $0$.

Now for the upper limit: $x=e^{37}$. Plugging this in, we get $t=\pi \ln (e^{37})$. Using the property of logarithms, the $37$ comes down, and $\ln (e)=1$, leaving us with $t=37\pi$. Just like that, the intimidating $e^{37}$ has vanished, replaced by a clean, manageable $37\pi$.

Now, our integral has transformed into:

This is no longer a complex logarithmic expression; it is the area under a simple sine wave. Visualize the sine curve. It oscillates between $1$ and $-1$, creating lobes above and below the axis.

Each full cycle, from $0$ to $2\pi$, consists of one positive lobe and one negative lobe. Because of the symmetry of the sine function, the area of the positive lobe is exactly $2$, and the area of the negative lobe is $-2$. When you add them, they cancel out to zero.

Since $37\pi$ contains exactly $18$ full periods (from $0$ to $36\pi$) and one final half-period (from $36\pi$ to $37\pi$), the first $36\pi$ of the integral contributes exactly zero to the total area.

We are left with only the final lobe:

Because the sine function is periodic, the area of this final lobe is identical to the area of the first lobe, ${\int }_0^{\pi }\sin (t)dt$.

Let us calculate this final piece: the antiderivative of $\sin (t)$ is $-\cos (t)$. Evaluating this from $0$ to $\pi$, we get:

The monster is defeated. The final answer is 2. You have navigated the complexity, respected the symmetry, and arrived at the truth. Keep this clarity with you as you face the next challenge.