Problem 111 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{0}^{k \pi} \sin \left[\frac{2 x}{\pi}\right] d x=A \cdot \frac{\sin k \sin \left(k+\frac{1}{2}\right)}{\sin \frac{1}{2}}\), where \(A\) is (A) \(\pi\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{2}\) (D) None of these

Step-by-Step Solution

Answer

A is \( \pi \).

Step 1: Understanding the Integral

We need to evaluate the definite integral \( \int_{0}^{k \pi} \sin \left[\frac{2 x}{\pi}\right] d x \). This integral involves the trigonometric function sine with a linear function of \(x\) inside.

Step 2: Substitution

To simplify the integral, perform a substitution. Let \( u = \frac{2x}{\pi} \). Then, \( du = \frac{2}{\pi} dx \), which means \( dx = \frac{\pi}{2} du \). The limits change as follows: when \( x = 0 \), \( u = 0 \), and when \( x = k\pi \), \( u = 2k \).

Step 3: Transforming the Integral

By substituting, we recast the integral as follows: \[ \int_{0}^{k \pi} \sin \left[\frac{2 x}{\pi}\right] dx = \int_{0}^{2k} \sin(u) \cdot \frac{\pi}{2} du. \]

Step 4: Evaluating the Integral

Evaluate the new integral: \[ \frac{\pi}{2} \int_{0}^{2k} \sin(u) du. \] The antiderivative of \( \sin(u) \) is \(-\cos(u)\). Thus, \[ \frac{\pi}{2} \left[ -\cos(u) \right]_{0}^{2k} = \frac{\pi}{2} \left[ -\cos(2k) + \cos(0) \right]. \]

Step 5: Simplifying the Result

Simplify the expression: \[ \frac{\pi}{2} [1 - \cos(2k)]. \] Using the identity \( 1 - \cos(2k) = 2\sin^2(k) \), it further simplifies to \( \pi \sin^2(k) \).

Step 6: Matching to the Given Expression

The expression can be further understood using the given formula: \( A \cdot \frac{\sin k \sin \left(k+\frac{1}{2}\right)}{\sin \frac{1}{2}} = \pi \sin^2(k) \). Simplifying the right side, we find that \( A = \pi \) when comparing both expressions.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify integrals. This method involves replacing a variable with a trigonometric function to make an integral more straightforward to evaluate. It is particularly useful when dealing with integrals involving square roots or specific trigonometric functions.
In the given exercise, we begin by performing a substitution. By letting \( u = \frac{2x}{\pi} \), we adjust our integral to take advantage of a simpler function. The derivative \( du = \frac{2}{\pi}dx \) informs us to change \( dx \) to \( \frac{\pi}{2}du \), which then allows us to seamlessly transform the integral limits. This step is crucial as it converts the original complicated trigonometric integral into a clearer form that is easier to solve. Through substitution, we're able to shift the focus from a composite sine function to a standard form that's more straightforward to handle.

Antiderivative

Finding an antiderivative, also known as indefinite integration, is the process of finding a function whose derivative yields the original function. In integrals involving trigonometric functions, knowing the standard antiderivatives is very helpful.
In our scenario, the integral \( \int \sin(u) \cdot \frac{\pi}{2} \, du \) requires us to find the antiderivative of \( \sin(u) \). The antiderivative of \( \sin(u) \) is \(-\cos(u)\), which provides a path to evaluate the definite integral once the limits are applied. By integrating and applying the original limits, we arrive at the evaluated definite integral \( \frac{\pi}{2} \left[ -\cos(2k) + \cos(0) \right] \). This evaluation relies on applying the correct antiderivative and ensures that we capture the area under the curve described by the original function within specified limits.

Trigonometric Identities

Trigonometric identities are essential mathematical tools that offer ways to express trigonometric functions in alternative forms. They simplify expressions and make complex problems more approachable. Some common identities include the Pythagorean identities, angle sum formulas, and double angle formulas.
In this exercise, the identity \( 1 - \cos(2k) = 2\sin^2(k) \) was used. This transformation is critical as it simplifies the expression \( \frac{\pi}{2} [1 - \cos(2k)] \) into \( \pi \sin^2(k) \). Such simplifications help in matching the final solved expression to the form given in the problem. By leveraging this identity, we can neatly wrap up the integral solution into the provided format, allowing an insight into its inherent nature and structure.