Animated Solution for Mathematics - Definite Integration: Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function such that $f(\frac{\pi }{4})=\sqrt{2},f(\frac{\pi }{2})=0$ and $f^{\prime }(\frac{\pi }{2})=1$ and let $g(x)={\int }_x^{\pi /4}(f^{\prime }(t)\sec t+\tan t\sec tf(t))dt$ for $x\in [\frac{\pi }{4},\frac{\pi }{2})$. Then ${\lim }_{x\to {\frac{\pi }{2}}^{-}}g(x)$ is equal to

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, might seem like a daunting calculus nightmare.

We are given a function $f(x)$ and an integral $g(x)$ defined as:

Our mission is to find the limit of $g(x)$ as $x$ approaches $\frac{\pi }{2}$ from the left. When you look at that integrand, $f^{\prime }(t)\sec t+f(t)\sec t\tan t$, do not panic. In the world of competitive mathematics, whenever you see a sum of two terms involving a function and its derivative, your intuition should immediately scream: Product Rule!

Let us recall the product rule from our toolkit: $\frac{d}{dt}(u\cdot v)=u^{\prime }v+u{v}^{\prime }$. Now, look at our integrand again.

If we let $u=f(t)$ and $v=\sec t$, then $u^{\prime }=f^{\prime }(t)$ and $v^{\prime }=\sec t\tan t$. The integrand is exactly $u^{\prime }v+u{v}^{\prime }$. This means the entire expression is just the derivative of the product $f(t)\sec t$.

This realization is the turning point of the problem. We have transformed a complex integral into the integral of a derivative, which is the most beautiful simplification in calculus.

By the Fundamental Theorem of Calculus, integrating a derivative returns the original function. So, our integral $g(x)$ becomes:

Evaluating this at the limits, we get:

We are given $f(\frac{\pi }{4})=\sqrt{2}$. Since $\sec (\frac{\pi }{4})=\sqrt{2}$, the first term is simply $\sqrt{2}\cdot \sqrt{2}=2$. Thus, our function simplifies to:

Now, we need to find ${\lim }_{x\to {\frac{\pi }{2}}^{-}}g(x)$. Substituting our expression, we have:

As $x\to {\frac{\pi }{2}}^{-}$, $f(x)\to f(\frac{\pi }{2})=0$ and $\cos x\to 0$. We have a $\frac{0}{0}$ indeterminate form! This is where we call upon L'Hopital's Rule.

We differentiate the numerator and denominator with respect to $x$:

Given $f^{\prime }(\frac{\pi }{2})=1$ and $\sin (\frac{\pi }{2})=1$, the limit becomes $\frac{1}{-1}=-1$. Finally, substituting this back into our expression for $g(x)$:

And there you have it! Through pattern recognition and the elegance of calculus, we have arrived at the answer: 3. Keep practicing, keep visualizing, and remember that every complex problem is just a collection of simple, beautiful truths waiting to be uncovered.