Animated Solution for Mathematics - Definite Integration: Let $\frac{d}{dx}F(x)=\frac{e^{\sin x}}{x},x>0$. If ${\int }_1^4\frac{2e^{\sin x^2}}{x}dx=F(k)-F(1)$, then one of the possible values of $k$ is $\dots \dots \dots$

Imagine you are standing at the edge of a complex mathematical landscape. You are given a derivative,

and asked to evaluate a seemingly daunting integral:

At first glance, this might look like a nightmare of transcendental functions, but let us take a breath. The Fundamental Theorem of Calculus is our master key here. It tells us that if $F^{\prime }(x)=f(x)$, then the integral of $f(x)$ is simply $F(x)$.

Look closely at the target integral:

The argument of the sine function is $x^2$, while the denominator is just $x$. This is a classic signal in calculus. We need the argument of the sine function to be a single variable to match our given derivative.

This screams for a substitution. We want to set $t=x^2$. But wait, if $t=x^2$, then $dt=2xdx$. We do not have a $2x$ in the numerator.

How do we fix this? We use a clever algebraic maneuver: multiply the integrand by $\frac{x}{x}$. This transforms our integral into:

Now, the $2x$ is waiting for us in the numerator, and the denominator has become $x^2$, which is just $t$.

Now, let us execute the substitution. With $t=x^2$, our differential $2xdx$ becomes $dt$. The denominator $x^2$ becomes $t$.

And most importantly, we must transform the limits of integration. When $x=1$, $t=1^2=1$. When $x=4$, $t=4^2=16$.

Our integral now becomes:

Look at this result. It is a perfect match for our given derivative relation $\frac{d}{dt}F(t)=\frac{e^{\sin t}}{t}$. By the Fundamental Theorem of Calculus, this integral is simply $F(16)-F(1)$.

The problem tells us that our integral is equal to $F(k)-F(1)$. We have just calculated that the integral is $F(16)-F(1)$.

By comparing these two expressions, the conclusion is immediate and elegant:

You see, the complexity of the function $F(x)$ was just a distraction. By understanding the structure of the integral and applying the right substitution, we peeled back the layers to reveal a simple, beautiful answer. Never be intimidated by the notation; look for the underlying symmetry, and the path will always reveal itself.