Problem 106 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{0}^{\sin ^{2} x} \sin ^{-1}(\sqrt{t}) d t+\int_{0}^{\cos ^{2} x} \cos ^{-1}(\sqrt{t}) d t\) is equal to (A) \(\frac{\pi}{4}\) (B) \(\frac{\pi}{6}\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The total value is \(\frac{\pi}{2}\), thus answer is (D) None of these.

Step 1: Understand the Problem

We need to evaluate the total integral defined by two definite integrals: \( \int_{0}^{\sin^{2} x} \sin^{-1}(\sqrt{t}) dt \) and \( \int_{0}^{\cos^{2} x} \cos^{-1}(\sqrt{t}) dt \). These integrals depend on actual expressions involving inverse trigonometric functions, and the limits are sub-functions of \( x \).

Step 2: Use a Substitution Method

For the integral \( \int_{0}^{\sin^{2} x} \sin^{-1}(\sqrt{t}) dt \), let us substitute \( t = u^2 \) such that \( dt = 2u du \) and \( \sqrt{t} = u \). A similar approach will be used for the second integral.

Step 3: Simplify Using Limits

For the substitution \( t = u^2 \):\[\int_{0}^{\sin^{2} x} \sin^{-1}(\sqrt{t}) dt = \int_{0}^{\sin x} \sin^{-1}(u) 2u du\]Set limits from 0 to \( \sin x \), allowing computation within simpler domains. Similarly, for \( \int_{0}^{\cos^{2} x} \cos^{-1}(\sqrt{t}) dt \), let \( t = v^2 \), leading to:\[\int_{0}^{\cos x} \cos^{-1}(v) 2v dv\]Set limits from 0 to \( \cos x \).

Step 4: Explore Symmetry in Integral Expressions

Observe the symmetry:\1. \( \int_{0}^{u} (\sin^{-1}(v) 2v) dv = \int_{0}^{u} \frac{d}{dt} [-v \sin^{-1}(v) + \sqrt{1-v^2}] dv \)\[ = -[v \sin^{-1}(v) + \sqrt{1-v^2}]|_{0}^{u}\]2. \( \int_{0}^{v} (\cos^{-1}(v) 2v) dv = \int_{0}^{v} \frac{d}{dt} [-v \cos^{-1}(v) + v \sqrt{1-v^2}] dv \)\[ = -[v \cos^{-1}(v) + v \sqrt{1-v^2}]|_{0}^{v}\].

Step 5: Calculate Definite Integrals

Calculate specific values:To evaluate for limits \( u = \sin x \) and\( v = \cos x \), calculate:\\[-[u \sin^{-1}(u) + \sqrt{1-u^2}] |_{0}^{\sin x}- [v \cos^{-1}(v) + \sqrt{1-v^2}]|_{0}^{\cos x}\]Evaluate both expressions to sum them.

Step 6: Topic Integration Result

Both integral parts can be seen as contributing constants where:\[\int_{0}^{1} \sin^{-1}(t) dt + \int_{0}^{1} \cos^{-1}(t) dt = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]From symmetry, adjust for bounds adjustment:Result simplifies to constant value for defined bounds. Integrals represent sum of inverse function representations.

Key Concepts

Definite Integration

Definite integration is a process of calculating the value of an integral with specific limits. In our exercise, we have two definite integrals:

• One with limits from 0 to \( \sin^2 x \)
• Another from 0 to \( \cos^2 x \)

Definite integrals can be seen as finding the area under the curve between two points on the x-axis represented by these limits.
When you evaluate a definite integral, the result is a number that represents the accumulated quantity over that interval.
This technique is useful for calculating total quantities, such as area, volume, or other accumulations.

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integration. It involves changing the variable of integration to transform the integral into an easier form.
In our problem, we use substitution to simplify the integrals involving inverse trigonometric functions.
For the first integral, we set \( t = u^2 \), making \( dt = 2u \, du \). This converts the integral into a form that is more manageable:

• \( \int_{0}^{\sin x} \sin^{-1}(u) 2u \, du \)

Similarly, for the second integral, we use \( t = v^2 \) and convert it into:

• \( \int_{0}^{\cos x} \cos^{-1}(v) 2v \, dv \)

This change of variables allows us to evaluate the integrals over simpler limits, which facilitates easier calculation.

Symmetry in Integrals

Symmetry in integrals can greatly simplify the computation process. Recognizing symmetry allows us to pair integrals or cancel terms, reducing the complexity of integration tasks.
In the exercise, examining symmetry between the parts of the integrals can indicate if they can combine to produce a constant result.
Let's observe the expressions:

• \( \int_{0}^{u} (\sin^{-1}(v) 2v) \, dv \)
• \( \int_{0}^{v} (\cos^{-1}(v) 2v) \, dv \)

By recognizing how the expressions mirror and complement each other, you can leverage symmetry to evaluate the integrals. This often involves realizing equivalent areas contribute to a constant, understanding behaviors over the complete interval from 0 to \( 1 \), leading to simplified evaluation.

Trigonometric Identities

Trigonometric identities play a crucial role in solving problems involving trigonometric functions, including inverse trigonometric functions. These identities provide relationships between angles and sides of triangles, which are very useful in integration.
For inverse functions like \( \sin^{-1} \) and \( \cos^{-1} \), knowing identities such as:

• \( \sin^2 x + \cos^2 x = 1 \)

helps in altering the limits of integration and simplifying expressions.
When evaluating integrals from \( 0 \) to \( \sin^2 x \) or \( \cos^2 x \), these identities help recharacterize the expressions at different stages, providing clarity through familiar relationships.
Recognizing how these identities simplify computations or transform expressions makes solving integrals involving inverse trigonometric functions more straightforward.