Animated Solution for Mathematics - Definite Integration: The solution for $x$ of the equation ${\int }_{\sqrt{2}}^x\frac{dt}{t\sqrt{t^2-1}}=\frac{\pi }{2}$ is

Welcome, future engineer. Today, we are going to peel back the layers of a seemingly simple integral equation. In the world of JEE Advanced, problems are rarely just about computation; they are about recognizing patterns and maintaining your composure when the result defies your initial expectations.

Recognizing the Form

We begin with the equation:

When you see an integrand like $f(t)=\frac{1}{t\sqrt{t^2-1}}$, your mind should immediately race to your toolkit of standard integrals. This is not a random collection of functions; it is the exact derivative of the inverse secant function.

Recall that the derivative of ${\sec }^{-1}t$ is $\frac{1}{|t|\sqrt{t^2-1}}$. Since our lower limit is $\sqrt{2}$, we are working in the positive domain, so we can confidently state that the antiderivative is ${\sec }^{-1}t$. This realization is the key that unlocks the entire problem.

The Fundamental Theorem of Calculus

Now that we have our antiderivative, we apply the Fundamental Theorem of Calculus. We evaluate the definite integral by taking the difference of the antiderivative at the upper and lower limits:

This step transforms our calculus problem into a trigonometric one. We are now tasked with evaluating ${\sec }^{-1}\sqrt{2}$.

Ask yourself: at what angle $\theta$ is $\sec \theta =\sqrt{2}$? Since $\sec \theta =\frac{1}{\cos \theta }$, this is equivalent to $\cos \theta =\frac{1}{\sqrt{2}}$. We know that $\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, so ${\sec }^{-1}\sqrt{2}=\frac{\pi }{4}$.

The Trigonometric Trap

Substituting this back into our equation, we get:

Isolating ${\sec }^{-1}x$, we add $\frac{\pi }{4}$ to both sides:

Here is where many students stumble. They assume $x$ must be positive, but look at the angle $\frac{3\pi }{4}$. This angle lies in the second quadrant.

In the second quadrant, the cosine function—and consequently the secant function—is negative. We must calculate $x=\sec (\frac{3\pi }{4})$. Since $\cos (\frac{3\pi }{4})=-\frac{1}{\sqrt{2}}$, it follows that $\sec (\frac{3\pi }{4})=-\sqrt{2}$.

Trusting Your Math

We have arrived at $x=-\sqrt{2}$. Now, we look at our options.

If you do not see $-\sqrt{2}$ listed, do not panic. In JEE Advanced, the option "None" is not a sign of failure; it is a test of your confidence.

You have performed the integration, applied the limits, navigated the trigonometric quadrants, and arrived at a mathematically sound result. If that result is not among the choices, then "None" is the correct answer. Stand tall in your derivation, trust your process, and move forward with the knowledge that you have mastered the logic behind the problem.