Problem 210 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution for \(x\) of the equation \(\quad\) [2007] \(\int_{\sqrt{2}}^{x} \frac{d t}{t \sqrt{t^{2}-1}}=\frac{\pi}{2}\) (A) 2 (B) \(\pi\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(-\sqrt{2}\)

Step-by-Step Solution

Answer

The solution is (D) \(-\sqrt{2}\).

Step 1: Understand the Problem

We need to find the value of \(x\) such that the integral \( \int_{\sqrt{2}}^{x} \frac{d t}{t \sqrt{t^{2}-1}} = \frac{\pi}{2} \) holds true. The given options are numerical or symbolic values, suggesting that \(x\) might be another common angle or identity in trigonometry/integrals.

Step 2: Recognize Integral Form

The integral \( \int \frac{dt}{t \sqrt{t^2 - 1}} \) is a standard form that can be solved using trigonometric substitution. It is related to the integral of the form \( \sec^{-1}(t) + C \), where the indefinite integral leads to the inverse hyperbolic functions.

Step 3: Solve the Indefinite Integral

From the standard integral form, we know that \( \int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + C \). Thus, our original definite integral \( \int_{\sqrt{2}}^{x} \frac{d t}{t \sqrt{t^{2}-1}} \) equals \( \sec^{-1}(x) - \sec^{-1}(\sqrt{2}) \).

Step 4: Evaluate Definite Integral

Since \( \sec^{-1}(\sqrt{2}) \) is \( \frac{\pi}{4} \) (because \( \sec(\frac{\pi}{4}) = \sqrt{2} \)), we set up the equation \( \sec^{-1}(x) - \frac{\pi}{4} = \frac{\pi}{2} \). Solving for \(x\), we have \( \sec^{-1}(x) = \frac{3\pi}{4} \).

Step 5: Solve for x

Equating \( \sec^{-1}(x) = \frac{3\pi}{4} \) to \( \sec(\frac{3\pi}{4}) \), we get \( x = \sec(\frac{3\pi}{4}) = -\sqrt{2} \). Hence, \(x = -\sqrt{2}\), which matches one of the given options.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a helpful technique in calculus to simplify integrals involving radicals.
When we come across an integral like \( \int \frac{dt}{t \sqrt{t^2 - 1}} \), it suggests using a substitution that can eliminate the square root. Here, the form resembles a textbook example that could benefit from trigonometric substitution, particularly using a substitution related to the secant function.

Why secant? Because the presence of \( t^2 - 1 \) fits the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \). Thus, we set \( t = \sec(\theta) \).

• This substitution helps transform \( t^2 - 1 \) into a manageable \( \tan^2(\theta) \).
• It also leads \( dt \) to become \( \sec(\theta)\tan(\theta)d\theta \).

After substituting, the integral becomes more straightforward to evaluate since it matches familiar trigonometric identities.

Inverse Hyperbolic Functions

Inverse hyperbolic functions may seem intimidating, but they are similar to inverse trigonometric functions.
In our problem, they emerge in the solution phase when we realize that a specific integral form can relate to inverse hyperbolic functions. When dealing with \( \int \frac{dt}{t \sqrt{t^2 - 1}} \), resolving it in terms of the arc secant function, denoted as \( \sec^{-1}(t) \), becomes intuitive.

What is the arc secant function? It’s the inverse of the secant function, and it tells us which angle, \( \theta \), corresponds to a given secant value.

• It helps solve integrals that revert back to standard forms involving inverse calculations.
• This is useful as it directly links integrals to the trigonometric angle solutions we are familiar with.

In our situation, identifying that \( \sec^{-1}(t) \) applies enabled us to move the problem from the integration world back to a trigonometric perspective, leading us to the solution \( x = -\sqrt{2} \).

Integration Using Standard Forms

Using standard forms for integration is like using a map on a familiar path.
Formulas for common integrals allow us to solve problems without going through the derivation each time.
One such formula in solving our exercise is \( \int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + C \). This correlates well to inverse functions and transforms definitively between trigonometric and algebraic expressions.

Why is this important?

• These forms are quick references to a known transformation and integration process.
• They prevent repetitive, tedious calculations and help you focus on applying the right technique quickly.

In our context, it allowed for the recognition of the problem’s structure and provided direct access to moving forward with the given bounds and target function forms. Efficiently using standard integration forms bridges problem identification and solution, letting us solve such exercises with minimal confusion and error.