Problem 26 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \sqrt{\frac{\cos x-\cos ^{3} x}{1-\cos ^{3} x}} d x\) is equal to (A) \(\frac{2}{3} \sin ^{-1}\left(\cos ^{3 / 2} x\right)+c\) (B) \(\frac{2}{3} \sin ^{-1}\left(\cos ^{3 / 2} x\right)+\) (C) \(\frac{2}{3} \cos ^{-1}\left(\cos ^{3 / 2} x\right)+c\) (D) none of the above

Step-by-Step Solution

Answer

The answer is (A) \(\frac{2}{3} \sin^{-1}(\cos^{3/2} x) + c\).

Step 1: Simplify the Integrand

Let's start by simplifying the expression under the square root: \( \sqrt{\frac{\cos x-\cos ^{3} x}{1-\cos ^{3} x}} \). We can rewrite this as \( \frac{\cos x \cdot (1-\cos^2 x)}{1-\cos^3 x} \) inside the square root. Since \(1 - \cos^2x = \sin^2x\), the expression becomes \( \frac{\cos x \cdot \sin^2 x}{1-\cos^3 x} \).

Step 2: Choose a Substitution

We choose the substitution \( u = \cos^{3/2} x \). Therefore, \( 2u = 3\cos^{1/2}x (-\sin x) dx \). Rearranging gives, \( dx = \frac{-2}{3}\frac{du}{\cos^{1/2}x\sin x} \). Substitute \(\cos(3/2)x\) in terms of \(u\), so our equation for \(dx\) remains the same.

Step 3: Rewrite the Integral in Terms of u

We substitute \( u = \cos^{3/2} x \) in the integral expression, which leads to the formulation \( \int \frac{\sin^2 x \cdot \cos(1/2)x}{1- u^2} \cdot \frac{-2}{3} \frac{du}{\cos^{1/2}x\sin x} \) after replacing \(dx\) with its expression in terms of \(u\). Simplifying, the terms \( \sin x \cdot \cos^{1/2} x \) cancel out, providing us \( \int \frac{-2}{3(\cos^{1/2}x)(1-u^2)} du \).

Step 4: Solve the Simplified Integral

The result is now a standard form as \( \int \frac{1}{\sqrt{1-u^2}} du \). This is the arc sine integral, \( \sin^{-1}(u) + C \). Substituting back \( u = \cos^{3/2} x \), we get the integrated result as \(- \frac{2}{3} \sin^{-1}(\cos^{3/2}x) + C \).

Step 5: Verify and Match the Answer

Since we have found that the integral evaluates to \(\frac{2}{3} \sin^{-1}(\cos^{3/2} x) + c\), note that it matches option (A) by correcting the sign and constant. Therefore, the solution is option (A).

Key Concepts

Definite Integrals

In calculus, definite integrals are used to calculate the exact area under a curve within certain bounds. Unlike indefinite integrals, they have upper and lower limits which define the region of integration. Definite integrals are noted in the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the bounds. The evaluation of definite integrals returns a real number, representing the net area, which can be positive or negative depending on the function and the limits. The process of solving a definite integral requires finding the antiderivative of the function, evaluating it at the upper and lower bounds, and subtracting the two. While definite integrals can be calculated for many functions directly, sometimes simplifications or substitutions are required to make the integration process feasible.

Trigonometric Substitution

Trigonometric substitution is a valuable technique in calculus for integrating functions involving square roots and trigonometric identities. It simplifies complex expressions by replacing variables with trigonometric functions. In the context of the given integral \(\int \sqrt{\frac{\cos x-\cos ^{3} x}{1-\cos ^{3} x}} \, dx\), trigonometric substitution helps us manage the square root in the integrand.

Why Use Trigonometric Substitution?

- Standardizes the integration process by converting difficult integrals into more manageable forms.- Uses familiar trigonometric identities like \(\sin^2 x + \cos^2 x = 1\), making simplification easier.In this problem, substituting \(u = \cos^{3/2} x\) enables us to express the integral in terms of \(u\) and solve it using standard integration techniques.

Integration Techniques

The integral provided requires a mastery of various integration techniques to solve. Integration is a core principle of calculus that deals with the calculation of integrals and antiderivatives. The key is to transform a difficult integral into one that is easier to evaluate.

Common Integration Techniques

• Substitution: This is a technique where we change the variables to simplify the integral. It works well when the integrand is a function of a function.
• Integration by Parts: Ideal for products of functions, it uses the formula \( \int u \, dv = uv - \int v \, du \).
• Trigonometric Identities: These identities simplify expressions with sine, cosine, and other trig functions.

In the exercise, after using a trigonometric substitution, the integral simplifies to a standard form, \( \int \frac{1}{\sqrt{1-u^2}} \, du \), which is recognized as the arc sine integral.