Question

Evaluate the following integrals: (i) \(\int \sqrt{2+\tan ^{2} x} d x\) (ii) \(\int \sqrt{1+\sec x} d x\) (iii) \(\int \sqrt{1+\operatorname{cosec} x} \mathrm{dx}\) (iv) \(\int \operatorname{cosec} x \sqrt{\cos 2 x} d x\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \sqrt{2+\tan ^{2} x} dx\) (ii) \(\int \sqrt{1+\sec x} dx\) (iii) \(\int \sqrt{1+\operatorname{cosec} x} dx\) (iv) \(\int \operatorname{cosec} x \sqrt{\cos 2 x} dx\) Answer: (i) \(\int \sqrt{2+\tan^2 x} dx = \ln(\sec x + \tan x) + C\) (ii) \(\int \sqrt{1+\sec x} dx = \cosh^{-1}(\sec x) + C\) (iii) \(\int \sqrt{1+\operatorname{cosec} x} dx = -\frac{1}{2}(\cos x \sqrt{1+\cos^2 x} + \sinh^{-1}(\cos x)) + C\) (iv) \(\int \operatorname{cosec} x \sqrt{\cos 2 x} dx = \frac{\sqrt{2}}{2} \left(\tan \left(\sec^{-1}\left(\frac{\cos x}{\sqrt{2}}\right)\right) + C\right)\)

Step-by-step solution

Rewrite the integrand using the trigonometric identity

Recall the identity \(\tan^2x = \sec^2x - 1\). So, we have: \(\int \sqrt{2+\tan^2 x} dx = \int \sqrt{1+ (\sec^2 x -1)} dx = \int \sqrt{\sec^2 x} dx\)

Integrate the expression

Now, the expression becomes: \(\int \sqrt{\sec^2 x} dx = \int \sec x dx\) Now, we know that the antiderivative of \(\sec x\) is \(\ln|\sec x + \tan x| + C = \ln(\sec x + \tan x) + C\), so our final answer is: \(\int \sqrt{2+\tan^2 x} dx = \ln(\sec x + \tan x) + C\) (ii) \(\int \sqrt{1+\sec x} d x\):

Use substitution

Let \(u = \sec x\), then \(du = \sec x \tan x dx\). We need to find an expression for \(dx\), so we can rewrite the integral in terms of \(u\). Dividing both sides by \(\sec x \tan x\), we get: \(dx = \frac{1}{\sec x \tan x} du\) Now, we rewrite the integral: \(\int \sqrt{1+\sec x} dx = \int \sqrt{1+u} \frac{1}{u \sqrt{u^2 - 1}} du\)

Simplify the integrand and integrate

The integrand simplifies to: \(\int \frac{1}{\sqrt{u^2 - 1}} du\) The antiderivative of \(\frac{1}{\sqrt{u^2 - 1}}\) is \(\cosh^{-1}(u) + C\). So, our final answer is: \(\int \sqrt{1+\sec x} dx = \cosh^{-1}(\sec x) + C\) (iii) \(\int \sqrt{1+\operatorname{cosec} x} \mathrm{dx}\):

Rewrite the integrand using the trigonometric identity

We know that \(\cosec x = \frac{1}{\sin x}\). So, the integral becomes: \(\int \sqrt{1+\frac{1}{\sin^2 x}} dx\)

Simplify the integrand

We can rewrite the expression under the square root as: \(\int \sqrt{\frac{\sin^2 x + 1}{\sin^2 x}} dx = \int \frac{\sqrt{1+\cos^2 x}}{\sin x} dx\)

Use substitution

Let \(u = \cos x\), then \(du = -\sin x dx\). We need to find an expression for \(dx\), so we can rewrite the integral in terms of \(u\). Dividing both sides by \(-\sin x\), we get: \(dx = -\frac{1}{\sin x} du\) Now, we rewrite the integral: \(\int \frac{\sqrt{1+\cos^2 x}}{\sin x} dx = -\int \sqrt{1+u^2} du\)

Integrating

The technique to compute the integral of this function is a bit more involved. It includes integrating by parts and using the functions \(\sinh x\) and \(\cosh x\) followed by back-substitution. The integral is given by: \(-\int \sqrt{1+u^2} du = -\frac{1}{2}(u \sqrt{1+u^2} + \sinh^{-1}(u)) + C\) Now, replace \(u\) with \(\cos x\): \(\int \sqrt{1+\operatorname{cosec} x} \mathrm{dx} = -\frac{1}{2}(\cos x \sqrt{1+\cos^2 x} + \sinh^{-1}(\cos x)) + C\) (iv) \(\int \operatorname{cosec} x \sqrt{\cos 2 x} d x\):

Rewrite the integrand using the double angle identity and trigonometric identity

We use the double angle identity \(\cos 2x = 2\cos^2 x - 1\), and the identity \(\cosec x = \frac{1}{\sin x}\). The integral becomes: \(\int \frac{1}{\sin x} \sqrt{2\cos^2 x - 1} dx\)

Use substitution

Let \(u = \cos x\), then \(du = -\sin x dx\). We need to find an expression for \(dx\), so we can rewrite the integral in terms of \(u\). Dividing both sides by \(-\sin x\), we get: \(dx = -\frac{1}{\sin x} du\) Now, we rewrite the integral: \(\int \frac{1}{\sin x} \sqrt{2\cos^2 x - 1} dx = -\int \sqrt{2u^2 - 1} du\)

Integrating

The integral can be solved with techniques like a trigonometric substitution, which may be a bit more involved. Let's substitute \(u = \frac{1}{\sqrt{2}} \sec \theta\) (wherewith a bit of work), leading to: \(\int \sqrt{2u^2 - 1} du = \frac{\sqrt{2}}{2} \int \sec^2 \theta d \theta = \frac{\sqrt{2}}{2} (\tan \theta + C)\)

Back-substitute and simplify the result

Now, replace \(u\) with \(\cos x\) and use the given substitution for \(u\). We get: \(\int \operatorname{cosec} x \sqrt{\cos 2 x} d x = \frac{\sqrt{2}}{2} \left(\tan \left(\sec^{-1}\left(\frac{\cos x}{\sqrt{2}}\right)\right) + C\right)\) This may not be the most simplified form, but it's the antiderivative for the given integral.

Key concepts

Integration Techniques

Integral calculus is a cornerstone of mathematics with applications ranging from physics to engineering. Mastering integration techniques is essential for success in calculus, especially for students preparing for competitive exams like the IIT JEE. One primary technique is substitution, where you replace a part of the integral with a new variable to simplify the expression. Another widely used method is integration by parts, which comes from the product rule for derivatives, and is crucial for integrals involving products of functions. Partial fraction decomposition is also an effective technique when dealing with rational functions.

For example, in IIT JEE problems, you often encounter complex trigonometric functions. Students can simplify these by using substitution or trigonometric identities. In the solution provided, notice how the complex integrands are simplified by rewriting them using identities, followed by a substitution to make integration straightforward. For each integral, an appropriate technique enables the simplification of the problem to reach the solution, and sometimes, a combination of methods is required.

• Use substitution to simplify the integrand and rewrite the integral in terms of a new variable.
• Apply integration by parts when faced with products of functions.
• Utilize partial fractions to break down complex rational functions.

Trigonometric Substitutions

Getting to grips with trigonometric substitutions can transform a difficult integral into a manageable one, a tactic particularly useful when integrating functions involving square roots. This substitution strategy often involves replacing expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\) with trigonometric functions that simplify the integrand.

Consider the example where \(\cosec x\) and \(\cos 2x\) appear under a square root. Recognizing the identity \(\cos 2x = 2\cos^2 x - 1\) is crucial before making a substitution. The substitution \(u = \cos x\) turns the expression into a form with \(u\), where familiar integration techniques can be applied. Understanding the relationship between the trigonometric functions allows students to deftly manipulate the integrand.

Benefits of Trigonometric Substitution

Using trigonometric substitution can:

• Make otherwise non-integrable expressions integrable.
• Simplify complex algebraic structures.
• Reveal underlying patterns and symmetries in the function.

Indefinite Integrals

Working through problems related to indefinite integrals, where there are no specified limits of integration, is a common practice among IIT JEE aspirants. The goal is to find the most general form of the antiderivative for a given function. Unlike definite integrals, the result of an indefinite integral includes a constant of integration \(C\), reflecting the infinite family of antiderivatives.

For instance, in evaluating the integral \(\int \sqrt{2+\tan^2 x} dx\), one must find the antiderivative of a square root of a trigonometric function. An indefinite integral teaches students the importance of recognizing core calculus principles and applying them effectively. It's a process of unraveling the mathematics to discover the function whose derivative corresponds to the integrand.

• Seek out the original function whose derivative matches the integrand.
• Include the constant of integration \(C\) in your answer.
• Remember that integral and derivative are inverse operations.

Trigonometric Identities

A proper understanding of trigonometric identities is critical when handling integrals involving trigonometric functions. These identities allow for the simplification of integrands, enabling the use of standard integration techniques. In the context of IIT JEE, identities such as \(\tan^2x = \sec^2x - 1\), \(\sin^2x + \cos^2x = 1\), and the double angle formula \(\cos 2x = 2\cos^2 x - 1\) are invaluable.

In our given example solutions, the smart use of these identities transforms the problem from a formidable challenge into an approachable task. Mastery of these identities is a powerful tool in the toolkit of any student, as they are crucial for evaluating many integrals that appear in calculus. Often, recognizing which identity to use is the key to solving an integral.

• Memorize fundamental trigonometric identities for simplification.
• Learn to identify which identity to use in a given problem.
• Combine them with other integration techniques for best results.