Problem 77 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{1}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral is \( \arcsin x + C \).

Step 1: Recognize the Form

The integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) resembles a standard trigonometric form. Specifically, it matches the derivative of the inverse sine function, \( \frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}} \).

Step 2: Apply the Antiderivative

Given that the integral matches the derivative of the \( \arcsin x \) function, we can directly write the antiderivative as \( \arcsin x \), plus a constant of integration \( C \).

Step 3: Write Down the General Solution

Thus, the indefinite integral of \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is \( \arcsin x + C \), where \( C \) is the integration constant.

Key Concepts

Trigonometric Integrals

Trigonometric integrals are a fascinating subset of integration that involve trigonometric functions like sine, cosine, tangent, and their relatives. When dealing with these, it's vital to know certain identities and derivatives of trigonometric functions. This knowledge can simplify the integration process significantly. In our given exercise, the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is a straightforward example where understanding these formulas makes it easier. Specifically, this integral resembles a known form connected to the derivative of an inverse trigonometric function. Recognizing these patterns is key when working with trigonometric integrals.One helpful tip is to keep a list of common trigonometric integrals handy. This includes integrals like:

• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)
• \( \int \sec^2 x \, dx = \tan x + C \)

These are the building blocks to tackle more complex trigonometric integrals.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function that reverses the process of differentiation. It represents a family of functions whose derivative is the given function. In simple terms, if you have a function \( f(x) \), the antiderivative of \( f(x) \) is another function, \( F(x) \), such that \( F'(x) = f(x) \). The process of finding an antiderivative is called integration, and when dealing with indefinite integrals, we include an arbitrary constant, usually denoted as \( C \). This constant arises because differentiation of a constant is zero, so antiderivatives aren't unique.In the context of our exercise, since the derivative of \( \arcsin x \) is \( \frac{1}{\sqrt{1-x^2}} \), it follows that the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) is simply \( \arcsin x \) plus a constant \( C \). This highlights the reverse nature of integration compared to differentiation.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions. They are crucial in calculus, especially for solving problems involving integrals and derivatives. The primary ones you will encounter are \( \arcsin x \), \( \arccos x \), and \( \arctan x \). These functions come into play when we need to "undo" a trigonometric function. For instance, \( \arcsin x \) represents the angle whose sine is \( x \). Therefore, it's the inverse operation of the sine function. In the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \), recognizing the expression as the derivative of \( \arcsin x \) helps us find its antiderivative directly. This way, learning the derivatives of inverse trigonometric functions results in a powerful tool for identifying particular forms in integrals. Keeping these derivatives at hand can save a lot of time in solving integrals involving inverse trigonometry. Whenever you face an integral containing an expression involving \( \sqrt{1-x^2} \), \( \sqrt{a^2-x^2} \), or similar terms, it might be beneficial to recall these inverse trigonometric relationships.