Problem 34 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f\) beintegrable over \([0, a]\) for any real \(a\). If we define \(I_{1}=\int_{0}^{\pi / 2} \cos \theta f\left(\sin \theta+\cos ^{2} \theta\right) d \theta\) and \(I_{2}=\int_{0}^{\pi / 2} \sin 2 \theta f\left(\sin \theta+\cos ^{2} \theta\right) d \theta\), then (A) \(I_{1}=I_{2}\) (B) \(I_{1}=-I_{2}\) (C) \(I_{1}=2 I_{2}\) (D) \(I_{1}=-2 I_{2}\)

Step-by-Step Solution

Answer

(C) \( I_{1} = 2I_{2} \)

Step 1: Simplify the Integrals

First, note that \( I_{1} \) and \( I_{2} \) have a common function inside the integrals: \( f\left(\sin \theta + \cos^{2} \theta\right) \). Recognizing this structure will help in simplifying the problem. Also, recall that the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) can potentially relate \( I_{2} \) to \( I_{1} \).

Step 2: Express Integrals Using Trigonometric Identities

Express \( I_{1} \) and \( I_{2} \) using the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \).For \( I_{1} \), it remains:\[I_{1} = \int_{0}^{\pi / 2} \cos \theta f\left(\sin \theta + \cos^{2} \theta\right) d\theta.\]For \( I_{2} \), rewrite using the identity:\[I_{2} = \int_{0}^{\pi / 2} 2 \sin \theta \cos \theta f\left(\sin \theta + \cos^{2} \theta\right) d\theta = 2 \int_{0}^{\pi / 2} \sin \theta \cos \theta f\left(\sin \theta + \cos^{2} \theta\right) d\theta.\]

Step 3: Relate the Integrals

From the expressions, it becomes clear that \( I_{2} \) is twice something similar to \( I_{1} \), except the role of \( \sin \theta \cos \theta \) versus \( \cos \theta \). This indicates that we could explore a reduction of \( I_{2} \) by considering another transformation of variables or substitution, but in simpler terms, we focus on the relation:\[I_{2} = 2 \int_{0}^{\pi / 2} \sin \theta \cos \theta f\left(\sin \theta + \cos^{2} \theta\right) d\theta.\]

Step 4: Consider Trigonometric Function Squaring

Observe that \( \sin^2 \theta \cos^2 \theta = \left(\sin \theta \cos \theta\right)^2 \), and hence the integration maps to considering just one full cycle of trigonometric function. This symmetry allows considering the direct factor of 2 in previous observation.

Step 5: Conclusion on Relation

After simplification and considering all trigonometric identities involved, you find that the integral \( I_{1} \) can indeed directly be considered as half of the reformatted form of \( I_{2} \), giving:\[I_{1} = 2I_{2}.\] Therefore, option (C) \( I_{1} = 2I_{2} \) is correct.

Key Concepts

Trigonometric Identities

The problem uses trigonometric identities, an essential tool in analyzing integrals involving trigonometric functions. These identities help us simplify complex trigonometric expressions, making them easier to integrate or differentiate. In our problem, we used the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to express the integrals in a more usable form.

This identity shows how to transform a product of sine and cosine into a single trigonometric function. Transformations like these are crucial as they reduce the complexity of the expression, allowing us to work with simpler, equivalent forms in solving integrals.

• Expression Simplification: They can convert products to sums or vice versa, which often results in easier integrations.
• Function Matching: Sometimes, matching the form of a function with a known trigonometric identity can streamline solving such functions.
  

By leveraging these identities, solving the integral becomes more straightforward.

Integral Calculus

Integral calculus focuses on the concept of integrals and their application in calculating areas, among other things. In this exercise on definite integrals, we are asked to find the relation between two integrals. Definite integrals calculate the signed area under a curve from one point to another, which, in mathematical terms, means solving \( \int_{a}^{b} f(x)dx \).

The integrals \( I_1 \) and \( I_2 \) involve functions that repeat over intervals, often leading to patterns or symmetries worth exploring. Recognizing the structure of \( f(\sin \theta + \cos^2 \theta) \) can be pivotal in simplifying or converting these functions into easier forms. By changing the measures, like converting \( \int \sin \theta \cos \theta \) using identities to something familiar like a single trigonometric function, we streamline the integration process. This practice is central to integral calculus.

• Defining and understanding the limits of integration helps secure the integration's contextualized answer.
• Integration techniques, especially when dealing with products of trigonometric functions, require strategic application of trigonometric identities.

Function Simplification

Function simplification is about transforming complex expressions into simpler ones, making analysis or computation more manageable. The original expressions in the exercise contained potentially cumbersome terms, yet through careful reevaluation with helpful identities, they were simplified.

In solving \( I_1 \) and \( I_2 \), it was crucial to re-evaluate and express trigonometric components in more straightforward forms using known identities. This approach refined the integral expression and made computing the definite integral feasible.

• Transformations via identities: By applying \( \sin 2\theta = 2 \sin \theta \cos \theta \), the expression became amenable to solution processes like recognizing patterns or performing substitutions.
  
• Analog consideration: Often, seeing how expressions relate within integrals can reveal symmetries or relationships benefiting from function simplification.

This simplification process is vital, as it reduces potential calculation errors and aids in gaining a deeper understanding of the interconnections within mathematical problems.