Problem 83 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{\left(\sin ^{-1} x\right)^{2} d x}{\sqrt{1-x^{2}}}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{(\sin^{-1} x)^3}{3} + C \).

Step 1: Understand the Integral

The given integral is \( \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} \, dx \). The \( \sin^{-1} x \) function is the inverse of the sine function. The expression under the square root, \( \sqrt{1-x^2} \), is associated with the derivative of \( \sin^{-1} x \). This means we may need to use integration by parts to solve it.

Step 2: Identify Parts for Integration By Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Let \( u = (\sin^{-1} x)^2 \) and \( dv = \frac{1}{\sqrt{1-x^2}} \, dx \). This choice leverages the fact that the derivative of \( \sin^{-1} x \) is simpler and known.

Step 3: Compute \( du \) and \( v \)

First, differentiate \( u \): \( du = 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} \, dx \). Then, integrate \( dv \) to find \( v \): since \( v = \sin^{-1} x \), we have \( v = \sin^{-1} x \) since \( dv = \frac{1}{\sqrt{1-x^2}} \, dx \) gives this integral result.

Step 4: Apply the Integration By Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). This results in \( \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} \, dx = (\sin^{-1} x)^2 \sin^{-1} x - \int \sin^{-1} x (2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}}) \, dx \). Simplify to get \( \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} \, dx = (\sin^{-1} x)^3 - 2\int (\sin^{-1} x)^2 \frac{1}{\sqrt{1-x^2}} \, dx \).

Step 5: Solve Final Integral

The integral \( \int (\sin^{-1} x)^2 \frac{1}{\sqrt{1-x^2}} \, dx \) repeats, leading to an equation we can solve algebraically. Let \( I = \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} \, dx \), then \( I = (\sin^{-1} x)^3 - 2I \), solve this as a simple algebraic equation for \( I \). Simplify to get \( 3I = (\sin^{-1} x)^3 \) and therefore \( I = \frac{(\sin^{-1} x)^3}{3} \). Increasing the constant of integration gives the solution.

Key Concepts

Understanding Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus as they help in solving integrals and differentiable functions involving trigonometric relationships. In this exercise, we are focusing on the inverse sine function, denoted as \( \sin^{-1} x \) or arcsin \( x \). This function is the inverse of the standard sine function, mapping angles back from sine values.

As with other trigonometric functions, \( \sin^{-1} x \) has specific differentiable properties. The most notable is its derivative, given by \( \frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}} \), which plays a crucial role in integration techniques.

• The derivative helps in formulating integrals, as seen here where it forms part of the denominator.
• Such functions are vital in simplifying and solving complex integrals, especially when combined with other techniques like integration by parts.

By understanding the differentiation properties of these inverse functions, we can better navigate integral calculus and its applications.

The Role of Integration By Parts in Integral Calculus

Integral calculus is one of the two main branches of calculus, the other being differential calculus. Integral calculus itself revolves around the concept of integration, which can be thought of as the reverse operation of differentiation.

In this exercise, we used the technique of integration by parts. This technique is akin to the product rule for differentiation and is applicable when we are integrating the product of two functions.

• The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).
• It requires us to identify two parts from the integrand: \( u \) and \( dv \), then determine their respective derivatives and integrals.

In this specific problem, we utilized integration by parts by setting \( u = (\sin^{-1} x)^2 \) and \( dv = \frac{1}{\sqrt{1-x^2}} \, dx \), facilitating a path to integrate functions that are product in nature.

Such techniques are fundamental in tackling integrals where direct integration is not straightforward.

Exploring Definite and Indefinite Integrals

Integrals in calculus can be classified as either definite or indefinite. Understanding their purpose and application helps in comprehending the broader concepts of integral calculus.

Definite Integrals:

• Represent the accumulation of quantities, such as area under a curve, bounded by limits of integration \( a \) to \( b \).
• Given by \( \int_{a}^{b} f(x) \, dx \), it results in a numerical value indicating total change.

Indefinite Integrals:

• Embody a family of functions represented by \( \int f(x) \, dx = F(x) + C \), where \( C \) is the integration constant.
• They describe the antiderivative or an original function from its derivative.

In this exercise, solving \( \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} \, dx \) represents an indefinite integral. It culminates in a general solution \( I = \frac{(\sin^{-1} x)^3}{3} + C \), emphasizing the capturing of all potential solutions without specific boundaries.

Both types of integrals are pivotal in linking differentiation and integration, portraying how quantities change over a given range.