Problem 141 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=\frac{\sin x}{4 \cos ^{4} x}+\frac{3 \sin x}{8 \cos ^{2} x}+\frac{3}{8} \ln \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} $$

Step-by-Step Solution

Answer

The exercise doesn't specify, but if it's supposed to find the integral, the integral of the given function is \( \frac{1}{4} \ln |\sec x| \). If it is to simplify the expression, it would be \( \frac{\sin x}{4} + \frac{3\sin x}{8} + \frac{3}{8} \ln \left|\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right| \).

Step 1 Perform Partial Fraction Decomposition

Look at the integral and determine what kind it is. I can tell this is not a plain power of x, or a simple trigonometric function, so I suspect it is a rational function. The numerator, \( \sin(x) \), is the derivative of the denominator, \( \cos(x) \), and this suggests a partial fraction decomposition may be involved.

Step 2 Apply Power Reduction to Denominators

For the fractions \( \frac{\sin x}{4 \cos ^{4} x} \) and \( \frac{3 \sin x}{8 \cos ^{2} x} \), replace \( \cos ^2 x \) with \( \frac{1 + \cos 2x}{2} \), and then replace \( \cos ^4 x \) with \( \frac{1 + \cos 2x}{2} \)^2.

Step 3 Simplify the Fractions

Now that all the denominators have been replaced with simpler functions, the fractions can be simplified. This allows us to simplify our integral into a form we can easily solve.

Step 4 Solve the Integral

Now the exercise is in a form that allows for a straightforward evaluation of the integral. The integral of each term individually corresponds to known formulas of elementary functions, which allows for immediate evaluation.

Step 5 Substitute Back the Original Variables

After finding the integral, remember to substitute the original variables back into your solution. This gives the solution in terms of the original question.

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental to understanding various mathematical problems, particularly those involving angles and periodic phenomena. In this exercise, the trigonometric function involved is the sine function, denoted as \( \sin x \), and the cosine function, \( \cos x \). These functions are critical in simplifying and solving integrals, especially when linked through identities or derivatives. The connection between \( \sin(x) \) and \( \cos(x) \) is pivotal; remember that \( \sin(x) = \cos(90^\circ - x) \), and vice versa for cosine.

Key trigonometric identities that often assist in manipulations include:

• Pythagorean Identity: \( \sin^2 x + \cos^2 x = 1 \).
• Angle Addition and Subtraction formulas, which let us break down complex angles into more manageable parts.
• Half-angle and double-angle formulas, such as \( \cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x \).

These identities allow us to simplify expressions involving trigonometric functions, like the ones found in the steps of the original solution.

Power Reduction Formula

Power reduction formulas are valuable for simplifying expressions with higher powers of trigonometric functions. They help in converting expressions to a form where integrals are easier to evaluate. In the original exercise, the power reduction formula used for cosine functions is crucial for simplifying fractions before integral evaluation.

The specific reduction formulas applied are:

• \( \cos^2 x = \frac{1 + \cos 2x}{2} \)
• \( \cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 \)

Using these forms makes the functions easier to integrate by eliminating higher powers of trigonometric expressions and replacing them with polynomial expressions of \( \cos(2x) \), which are more straightforward to integrate. This technique is especially handy in calculus, where simplifying a trigonometric integral or differential equation is needed for further computation.

Integral Evaluation

Integral evaluation is the process of finding the integral of a function, which represents the area under the curve of the graph of the function. In this exercise, the goal is to evaluate the integral after using techniques such as partial fraction decomposition and power reduction to simplify the expressions within the integral.

The integration process involves resolving the decomposed partial fractions into a sum of simpler integrals. Once the fractions are simplified using the power reduction formulas, they can be integrated individually using standard integral formulas. These known forms usually come from basic integral tables or standard calculus results involving polynomial, exponential, or trigonometric functions.

After calculating each integral separately, the last step involves combining these results and substituting back any variables that were replaced during simplification. This substitution step ensures that the answer relates to the original variables in the problem, completing the integral evaluation successfully.