Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of a curve passing through origin is given by \(y=\int x^{3} \cos x^{4} d x .\) If the equation of the curve is written in the form \(x=g(y)\), then (A) \(g(y)=\sqrt[3]{\sin ^{-1}(4 y)}\) (B) \(g(y)=\sqrt{\sin ^{-1}(4 y)}\) (C) \(g(y)=\sqrt[4]{\sin ^{-1}(4 y)}\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (C) \(g(y)=\sqrt[4]{\sin^{-1}(4y)}\).

Step 1: Determine the Integral

Let's first find the expression for the integral \( y = \int x^3 \cos(x^4) \, dx \). To evaluate this integral, use the substitution method. Set \( u = x^4 \), then \( du = 4x^3 \, dx \), or \( x^3 \, dx = \frac{du}{4} \). The integral becomes: \[ y = \int \cos(u) \cdot \frac{du}{4} = \frac{1}{4} \int \cos(u) \, du \].

Step 2: Evaluate the Integral

Now, evaluate \( \int \cos(u) \, du \), which is a standard integral. The antiderivative of \( \cos(u) \) is \( \sin(u) \). Thus, the complete integral for \( y \) is: \[ y = \frac{1}{4} \sin(u) + C \]. Given that the curve passes through the origin \((0,0)\), substitute \( x=0 \) and \( y=0 \) to find \( C \): \[ 0 = \frac{1}{4} \sin(0) + C \Rightarrow C = 0 \]. Therefore, \( y = \frac{1}{4} \sin(x^4) \).

Step 3: Express x in Terms of y

Now rewrite the equation by expressing \( x \) in terms of \( y \). Start with \( y = \frac{1}{4} \sin(x^4) \). To isolate \( x \), first multiply both sides by 4: \( 4y = \sin(x^4) \). Next, take the inverse sine of both sides: \( x^4 = \sin^{-1}(4y) \). Finally, solve for \( x \) by taking the fourth root: \( x = \sqrt[4]{\sin^{-1}(4y)} \).

Step 4: Identify g(y)

We've found that \( x = \sqrt[4]{\sin^{-1}(4y)} \). This expression matches option (C) perfectly, so \( g(y) = \sqrt[4]{\sin^{-1}(4y)} \).

Key Concepts

Substitution Method

The Substitution Method is an integral calculus technique used to simplify the process of finding the integral of complex functions. It's especially useful when dealing with integrals that involve composite functions.

In the given exercise, we applied the substitution method where we had the integral expression \( \int x^3 \cos(x^4) \, dx \). To make this integral easier to solve, we used the substitution \( u = x^4 \), which transformed the original expression into a simpler form.

Here's why substitution is advantageous:

• It simplifies complex integral expressions by changing the variable, making integration more straightforward and manageable.
• It often converts tricky integrals into basic forms that have known antiderivatives, such as \( \int \cos(u) \, du \), which is simply \( \sin(u) \).
• It's especially powerful for integrals involving compositions of functions, where the inner function's derivative is present elsewhere in the expression.

The key steps involve selecting an appropriate substitution (in our case \( u = x^4 \)), computing the derivative \( du \), and rewriting the integral in terms of \( u \). This method turns a complex integration problem into a manageable one.

Trigonometric Integrals

Trigonometric integrals are integrals that involve trigonometric functions like sine and cosine. These types of integrals are common in calculus and often require a clever approach for evaluation.

In the exercise, after using the substitution method to transform \( \int x^3 \cos(x^4) \, dx \) into \( \frac{1}{4} \int \cos(u) \, du \), the trigonometric aspect comes into play. The integral \( \int \cos(u) \, du \) is a classic trigonometric integral, whose antiderivative is known to be \( \sin(u) \).

Key aspects to remember about trigonometric integrals include:

• They often involve recognizing patterns in the integrand that relate to standard derivatives of sine, cosine, tangent, etc.
• Knowledge of basic trigonometric identities and derivatives simplifies the process, allowing transformation into solvable expressions.
• Applying integration techniques like substitution or integration by parts might assist in tackling more complex trigonometric integrals.

Successful evaluation of these integrals often requires a clear understanding of the fundamental relationships and properties of trigonometric functions.

Inverse Trigonometric Functions

Inverse Trigonometric Functions emerge when you solve equations that involve trigonometric identities. These functions, such as \( \sin^{-1}(x) \), help in determining angles when given their sine, cosine, or tangent values.

In our example, after obtaining \( y = \frac{1}{4} \sin(x^4) \), expressing \( x \) in terms of \( y \) required the use of inverse trigonometric functions. Solving \( 4y = \sin(x^4) \), we used the inverse sine to find \( x^4 = \sin^{-1}(4y) \).

Key insights into inverse trigonometric functions include:

• They are the reverse process of trigonometric functions, converting a ratio back to an angle.
• They're essential in solving calculus equations that require angle determination from trigonometric expressions.
• Inverse trigonometric functions appear frequently in integration, especially in problems involving arcs and angles.

Understanding these functions helps resolve equations that necessitate angle calculations. In calculus, they are integral to expressing variables in terms of others, as demonstrated in this exercise.