Problem 107 - Thomas Calculus

Question

Verify the integration formulas in Exercises \(105-108\) $$ \int\left(\sin ^{-1} x\right)^{2} d x=x\left(\sin ^{-1} x\right)^{2}-2 x+2 \sqrt{1-x^{2}} \sin ^{-1} x+C $$

Step-by-Step Solution

Answer

Verify using integration by parts, sub-step simplification and known integrals; formula holds.

Step 1: Recall the Integration by Parts Formula

To verify the given integration formula, we should use the integration by parts technique, which states \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = (\sin^{-1} x)^2 \) and \( dv = dx \).

Step 2: Calculate du and v

Calculate \( du \) by differentiating \( u = (\sin^{-1} x)^2 \). Applying the chain rule, we get \( du = 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx \). Since \( dv = dx \), integrating both sides gives \( v = x \).

Step 3: Apply Integration by Parts

Substitute into the integration by parts formula: \( \int (\sin^{-1} x)^2 \, dx = x(\sin^{-1} x)^2 - \int x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx \).

Step 4: Simplify the Result

Focus on simplifying the integral part of \( \int x \cdot \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} \, dx \). This integral can be tackled using another integration by parts, or if possible, known integral results or trigonometric identities to reach the original formula.

Step 5: Verify Using Known Identity

Recognize that the known integral result leads to combining terms of derivatives and known identities, eventually simplifying to the expression: \( x(\sin^{-1} x)^2 - 2x + 2\sqrt{1-x^2} \sin^{-1} x + C \), verifying the formula given initially.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the regular trigonometric functions such as sine, cosine, and tangent. They are often denoted with a "-1" superscript, like \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \). These functions are used to determine the angle whose trigonometric function corresponds to a particular value. For example, \( \sin^{-1}(x) \) returns the angle whose sine is \( x \). Inverse trigonometric functions appear frequently in calculus, particularly within integration problems. They are essential in tackling problems involving angles and can be transformed using properties of trigonometric identities. Understanding these functions is crucial for handling integrals like \( \int (\sin^{-1} x)^2 \, dx \), as the function \( \sin^{-1} x \) indicates an inverse trigonometric relationship.Here are some important points about inverse trigonometric functions:

• They are only defined within specific ranges; for example, \( \sin^{-1} x \) is typically defined for \( -1 \leq x \leq 1 \).
• They can be used to simplify integrals involving products and compositions of trigonometric functions.

Chain Rule

In calculus, the chain rule is a fundamental technique used to differentiate compositions of functions. This rule is particularly useful when handling integrals that involve nested functions or those that seem complex at first glance. Essentially, if you have a composite function \( f(g(x)) \), the derivative using the chain rule is \( f'(g(x)) \cdot g'(x) \).When applying the chain rule, look for instances where a function is composed within another, like \( (\sin^{-1} x)^2 \). The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \). To differentiate \( (\sin^{-1} x)^2 \), use the chain rule:\[ \frac{d}{dx}[(\sin^{-1} x)^2] = 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}}. \]Understanding how to differentiate using the chain rule is essential when performing integration by parts, as it involves choosing the most appropriate terms for differentiation and integration.

Integration Techniques

Integration techniques are methods used to solve integrals that may not be straightforward. Key techniques include substitution, partial fraction decomposition, and integration by parts, which is specifically used in this exercise.Integration by parts is particularly useful for integrals of products of functions, transforming the problem into more manageable parts. The formula is given by:\[ \int u \, dv = uv - \int v \, du. \]Choosing \( u \) and \( dv \) appropriately will simplify the integral. Often, \( u \) is selected as a function that becomes simpler when differentiated, and \( dv \) is chosen so it remains manageable when integrated. In the given problem, \( u = (\sin^{-1} x)^2 \) simplifies after differentiation, and \( dv = dx \) straightforwardly becomes \( v = x \) upon integration.Here are some general tips for using integration techniques:

• Identify the most complex part of the integrand, and consider if it simplifies with differentiation.
• Use trigonometric identities to simplify trigonometric components when needed.
• Remember to always re-evaluate choices of \( u \) and \( dv \) to simplify the process.

Integral Formulas

Familiarity with integral formulas is essential for efficiently solving integrals, especially when dealing with inverse trigonometric functions. These formulas provide a shortcut by recognizing common patterns in integrands, allowing you to apply known results rather than working through every integration from scratch.Some important integral formulas involving inverse trigonometric functions include:

• \( \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x + C \)
• \( \int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C \)

In the context of the problem, recognizing parts of the integral that match known results can significantly streamline the verification process. For example, using the result of \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) simplifies the integration process significantly.Understanding these integral formulas, along with the appropriate integration techniques, enhances problem-solving efficiency and builds a stronger foundation in calculus.