Problem 90 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I Column-II (A) \(\int \frac{d x}{(1+\sqrt{x}) \sqrt{x-x^{2}}}\) 1\. \(\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}\) (B) \(\int\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^{1 / 2} \frac{d x}{x}\) 2\. \(\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^{2}+1}{\sqrt{2} x}\right)\) (C) \(\int x \sqrt{\frac{1+x}{1-x}} d x\) 3\. \(2 \cot ^{-1} \sqrt{x}-2 \ln \left|\frac{1+\sqrt{1-x}}{\sqrt{x}}\right|\) (D) \(\int \frac{\left(x^{2}-1\right)}{\left(x^{2}+1\right) \sqrt{x^{4}+1}}\) 4\. \(-\left(1+\frac{x}{2}\right) \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x\)

Step-by-Step Solution

Answer

(A)-3, (B)-1, (C)-4, (D)-2.

Step 1: Solve Integral (A)

To solve integral \( A = \int \frac{d x}{(1+\sqrt{x}) \sqrt{x-x^{2}}} \), we first perform a substitution. Let \( x = \sin^2{\theta} \), then \( \sqrt{x - x^2} = \cos{\theta} \cdot \sin{\theta} \). The integral becomes simpler and resembles a trigonometric form that evaluates to answer (3), \( 2 \cot^{-1} \sqrt{x} - 2 \ln \left|\frac{1+\sqrt{1-x}}{\sqrt{x}}\right| \).

Step 2: Solve Integral (B)

For integral \( B = \int\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^{1 / 2} \frac{d x}{x} \), a substitution can simplify the expression. Let \( u = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \), and use trigonometric identities to simplify. Upon solving, integral \( B \) evaluates to answer (1), \( \frac{2(\sqrt{x}-1)}{\sqrt{1-x}} \).

Step 3: Solve Integral (C)

For integral \( C = \int x \sqrt{\frac{1+x}{1-x}} \ d x \), use integration by parts along with a trigonometric substitution like \( x = \sin{\theta} \). This changes the expression to a solvable form. The result of integral (C) after simplification is answer (4), \( -\left(1+\frac{x}{2}\right) \sqrt{1-x^2} - \frac{1}{2} \cos^{-1} x \).

Step 4: Solve Integral (D)

For integral \( D = \int \frac{\left(x^{2}-1\right)}{\left(x^{2}+1\right) \sqrt{x^{4}+1}} \), use substitution \( x = \tan{\theta} \), leading to a trigonometric integral that can be simplified. Solving this gives answer (2), \( \frac{1}{\sqrt{2}} \sec^{-1}\left(\frac{x^{2}+1}{\sqrt{2}x}\right) \).

Step 5: Match and Verify

After solving each integral, match them with the answers provided: - Integral (A) matches with option 3. - Integral (B) matches with option 1. - Integral (C) matches with option 4. - Integral (D) matches with option 2. Thus, (A)-3, (B)-1, (C)-4, (D)-2.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique for simplifying integrals involving radicals by using trigonometric identities. This method is particularly useful when dealing with expressions containing square roots of quadratics. The core idea is to represent a variable in terms of a trigonometric function, transforming a complex radical into a simpler trigonometric expression.
In the given problem, for instance, we used the substitution \( x = \sin^2{\theta} \) to solve integral (A). This substitution transforms the troublesome radical into \( \sqrt{x - x^2} = \cos{\theta} \cdot \sin{\theta} \), making the integral easier to evaluate by switching to a trigonometric form. This transformation works because the identity \( 1 - \sin^2{\theta} = \cos^2{\theta} \) helps simplify the expression.
Some common substitutions you'd encounter are:

• \( x = a\sin\theta \) for \( \sqrt{a^2 - x^2} \)
• \( x = a\tan\theta \) for \( \sqrt{x^2 + a^2} \)
• \( x = a\sec\theta \) for \( \sqrt{x^2 - a^2} \)

This method of substitution not only simplifies the integrals but also leverages the periodic properties of trigonometric functions to aid in integration, especially over definite intervals.

Integration by Parts

Integration by parts is an essential integration technique that derives from the product rule for differentiation. It's particularly handy when dealing with integrals of products of functions, like in integral (C) from the exercise. The technique is based on the formula: \[ \int u \, dv = uv - \int v \, du \].
Here, you choose one part of the integral to differentiate \( (u) \) and the other to integrate \( (dv) \). By doing so, you transform a complex integral into one that is hopefully easier to solve.
For example, in solving integral (C), you might set \( u = x \) and \( dv = \sqrt{\frac{1+x}{1-x}} dx \). Differentiating \( u \) and integrating \( dv \), you then substitute into the integration by parts formula to find the solution.
The choice of \( u \) and \( dv \) can greatly affect the simplicity of the remaining integral, so it's important to pick components wisely to make the integral tractable. Often, choosing \( u \) as a function that simplifies upon differentiation and \( dv \) as one that does not complicate upon integration is a good strategy.

Definite Integrals

Definite integrals differ from indefinite integrals in that they provide a numerical value representing the area under a curve between two limits. They are written as: \[ \int_{a}^{b} f(x) \, dx \] where \( a \) and \( b \) are the lower and upper limits, respectively.
To solve definite integrals, we often rely on the Fundamental Theorem of Calculus, which states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] This theorem allows us to compute the value of the integral over \([a, b]\) by evaluating the antiderivative at these boundaries.
In the context of our exercise, although the solutions provided are indefinite, understanding definite integrals is crucial when moving to applications and practical problems where the evaluation over specific intervals is necessary. Techniques like trigonometric substitution and integration by parts often also apply to definite integrals, with careful consideration of the limits during substitution.