Problem 4 - Calculus: Early Transcendentals

Question

Verify that \(\int 2 x \arccos x d x=x^{2} \arccos x-\frac{x \sqrt{1-x^{2}}}{2}+\frac{\arcsin x}{2}+C\)

Step-by-Step Solution

Answer

The given integral expression is verified to be correct.

Step 1: Understand the Problem

We are given the integral \( \int 2x \arccos x \, dx \) and need to verify whether it equals the expression \( x^{2} \arccos x - \frac{x \sqrt{1-x^2}}{2} + \frac{\arcsin x}{2} + C \). This involves differentiation to see if the derivative of the given expression equals the integrand.

Step 2: Differentiate the Expression

Let \( F(x) = x^{2} \arccos x - \frac{x \sqrt{1-x^2}}{2} + \frac{\arcsin x}{2} \). We need to calculate \( F'(x) \) and verify if it equals \( 2x \arccos x \).

Step 3: Use the Product Rule on \( x^2 \arccos x \)

The derivative of \( x^2 \arccos x \) using the product rule is \( 2x \arccos x + x^2 \frac{-1}{\sqrt{1-x^2}} \).

Step 4: Differentiate Other Terms

- The derivative of \( -\frac{x \sqrt{1-x^2}}{2} \) is found using the product rule: \(-\frac{1}{2}(\sqrt{1-x^2} - x \cdot \frac{-x}{\sqrt{1-x^2}}) = -\frac{\sqrt{1-x^2}}{2} + \frac{x^2}{2\sqrt{1-x^2}}\).- The derivative of \( \frac{\arcsin x}{2} \) is \( \frac{1}{2\sqrt{1-x^2}} \).

Step 5: Combine the Derivatives

Combine all derivatives:\[2x \arccos x + x^2 \frac{-1}{\sqrt{1-x^2}} - \frac{\sqrt{1-x^2}}{2} + \frac{x^2}{2\sqrt{1-x^2}} + \frac{1}{2\sqrt{1-x^2}}\]

Step 6: Simplify the Combined Derivative

Simplify the expression to see if it equals \( 2x \arccos x \):\[2x \arccos x - x^2 \frac{1}{\sqrt{1-x^2}} - \frac{\sqrt{1-x^2}}{2} + \frac{x^2+1}{2\sqrt{1-x^2}}\]This simplifies to \( 2x \arccos x \), confirming the initial integral calculation.

Key Concepts

Integration by Parts

Integration by Parts is a valuable method in calculus for finding integrals of products of functions. Imagine trying to integrate a function that's a product of two simpler functions, like \( f(x) \) and \( g(x) \). The formula to apply this technique is:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) is a function you choose to differentiate, and \( dv \) is the rest that you integrate. This approach ingeniously shifts the integration task from one we struggle with to one that's more manageable. When deciding what to choose for \( u \) and \( dv \), a common guideline is the ILATE rule:

• Inverse functions
• Logarithmic functions
• Algebraic functions
• Trigonometric functions
• Exponential functions

By prioritizing what to differentiate and integrate, we minimize complexity. In our given problem, the approach guided us towards simplifying the underlying expression before verifying the result.

Differentiation

Differentiation is the process of finding the derivative of a function, which is essentially the function's rate of change. In calculus, it's akin to looking at how a little alteration in \( x \) influences the value of \( y \), i.e., \( f'(x) \). For the given problem, differentiation plays a critical role in verifying the result of an integral.

When differentiating complex expressions, it’s often necessary to apply rules like the product rule or chain rule. These help tackle terms where functions are intertwined with each other. In our exercise, we used differentiation to verify the expression by checking that the derivative of the right side aligned with the integrand on the left.

By breaking down each term methodically, we ensure the derivative we derive matches the original expression precisely, validating the integral transformation.

Product Rule

The Product Rule is an essential tool for differentiation, particularly when dealing with the derivative of a product of two functions. If you have two functions \( u(x) \) and \( v(x) \), the product rule states:

• \( (uv)' = u'v + uv' \)

This rule shows how the changes in the individual functions contribute to changes in their product.

In the provided problem, the product rule was used to differentiate terms like \( x^2 \arccos x \). By applying the rule, we determined how the calculus principles apply to complex expressions involving multiple layers of functions. This approach allowed us to focus on each component's behavior, helping us simplify and verify the expression correctly.

Overall, mastering the product rule expands your toolkit for solving intricate calculus problems involving multiple function interactions.

Trigonometric Functions

Trigonometric Functions, such as \( \arccos(x) \) and \( \arcsin(x) \), are fundamental tools in calculus, vital for handling problems involving angles and lengths in mathematical expressions. These functions provide insight into periodic behaviors and geometric properties.

In our exercise, the functions \( \arccos(x) \) and \( \arcsin(x) \) played key roles in the verification process. Understanding these inverse trigonometric functions means recognizing that they map ratios back to angles, thus needing careful handling in differentiation and integration.

• For \( \arccos(x) \), its derivative is \( -\frac{1}{\sqrt{1-x^2}} \).
• For \( \arcsin(x) \), its derivative is \( \frac{1}{\sqrt{1-x^2}} \).

These derivatives are instrumental in confirming if integration and differentiation tasks are correctly executed and ensure the function’s behavior during problem-solving aligns with trigonometric identities. Approaching such functions with the right understanding simplifies their use in various calculus contexts.