Problem 11 - Calculus 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Integral of the Form (a) \(\int \frac{d x}{a+b \cos x}\) (b) \(\int \frac{d x}{a+b \sin x}\) (c) \(\int \frac{d x}{a+b \cos x+c \sin x}\) Working Rule (i) \(\mathrm{P}\) ut \(\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\) and \(\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\) so that the given integrand becomes a function of \(\tan \frac{x}{2}\) (ii) Put \(\tan \frac{x}{2}=t=>\frac{1}{2} \operatorname{scc}^{2} \frac{x}{2} d x=d t\) (iii) Integrate the resulting rational algebraic lunction of \(t\) (iv) In the answer, put \(t=\tan \frac{x}{2}\).

Step-by-Step Solution

Answer

Use the given trigonometric identities and substitutions to express each integrand in terms of \( t = \tan \frac{x}{2} \), simplify, integrate, and back-substitute for \( x \).

Step 1: Substitute Trigonometric Functions

We begin by substituting the trigonometric functions using the given identities. For (a) and (b), we have \( \cos x=\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} \) and \( \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}} \). For (c), both substitutions are used in the expression \( a+b \cos x+c \sin x \) so that it becomes a function of \( \tan \frac{x}{2} \).

Step 2: Substitute for dx

Now, let \( \tan \frac{x}{2} = t \), which implies that \( \frac{1}{2} \sec^2 \frac{x}{2} \, dx = dt \). Solve for \( dx \) the expression \( dx = \frac{2}{1 + t^2} \, dt \).

Step 3: Rewrite the Integrands

Rewrite each integrand in terms of \( t \):- For (a), substitute and simplify \( \int \frac{2 \, dt}{(a + \frac{b - bt^2}{1 + t^2})(1 + t^2)} \).- For (b), substitute and simplify \( \int \frac{2 \, dt}{a + \frac{2bt}{1 + t^2}}(1 + t^2) \).- For (c), simplify \( \int \frac{2 \, dt}{a + \frac{b - bt^2}{1 + t^2} + \frac{2ct}{1 + t^2}}(1 + t^2) \).

Step 4: Simplify and Integrate

Simplify each expression using algebraic manipulation to reduce the integrand into a form that can be further easily integrated with respect to \( t \). For each expression, compute the resulting integrals. Often, these will turn into a standard integral form involving \( t \).

Step 5: Back-substitute for x

After obtaining the integral in terms of \( t \), substitute back using \( t = \tan \frac{x}{2} \) to express your solution in terms of \( x \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in integral calculus. It is especially useful when dealing with integrals involving expressions like \( a+b\cos x\) or \( a+b\sin x\). Understanding how to manipulate these expressions can provide a direct path to solving complex integrals.

The primary idea is to express trigonometric functions such as \( \cos x\) and \( \sin x\) in terms of a new variable \( t = \tan \frac{x}{2} \). This substitution leverages the Weierstrass substitution, transforming complex trigonometric integrals into simpler rational functions of \( t \). This technique simplifies the integration process significantly.

• \( \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)
• \( \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)

By making these substitutions, the integral becomes a function of \( t \) instead of \( x \), allowing for more straightforward integration processes. Once integrated, we can then substitute back to express the final answer in terms of \( x \).

Integral Calculus

Integral calculus is a fundamental area of mathematics. Its main objective is to find the antiderivative, which is the reverse process of differentiation. When dealing with complex integrands, such as those involving trigonometric functions, special techniques like trigonometric substitution can be employed.

In this particular problem, the goal is to solve integrals of the form \( \int \frac{d x}{a+b\cos x} \), \( \int \frac{d x}{a+b\sin x}\), and \( \int \frac{d x}{a+b\cos x+c\sin x}\). These require substitution techniques that convert trigonometric expressions into rational ones in terms of a new variable \( t \). This is a classic example of how substitutions can simplify the integration process.

Once the substitution is made and the integrand is simplified, we can apply standard integration techniques available for rational functions. Finally, back-substitution returns the solution to its original variable form.

Rational Algebraic Functions

Rational algebraic functions appear frequently in calculus, especially when simplifying trigonometric integrals. In this exercise, after performing the trigonometric substitution, the given integrals become rational functions involving the variable \( t = \tan \frac{x}{2} \).

A rational function is defined as a quotient of two polynomials. The algebraic approach allows these functions to be manipulated easily, applying basic algebraic manipulations such as factoring, expanding, or simplifying to reduce the expression into a form that can be directly integrated.

For example, when the substitution is applied, the integrands are transformed and simplified to forms like \[ \int \frac{2 \, dt}{(a + \frac{b - bt^2}{1 + t^2})(1 + t^2)} \] or similar rational expressions. Integrating these forms involves identifying possible partial fractions or recognizing standard integral forms, facilitating the computational process.

Trigonometric Identities

Trigonometric identities are essential tools in calculus, particularly when simplifying expressions for easier manipulation and integration. In the context of this problem, we rely on identities that express \( \sin x \) and \( \cos x \) using \( \tan \frac{x}{2} \). This change of variables is pivotal in transforming the original problem into a more manageable form.

Some frequently used identities in this process include:

• \( \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)
• \( \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)

These identities help eliminate trigonometric functions, turning them into rational expressions. By applying these identities, we reduce the complexity of the integration process and simplify the algebraic manipulation. This transformation makes otherwise difficult integrals more accessible, laying the groundwork for straightforward solutions.