Problem 78 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{5}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The solution is \( 5 \sin^{-1}(x) + C \).

Step 1: Identify the integrand

The given integral is \( \int \frac{5}{\sqrt{1-x^2}} \, dx \). We notice that the integrand \( \frac{1}{\sqrt{1-x^2}} \) resembles the derivative of an inverse trigonometric function.

Step 2: Recall the Standard Result

We know the derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \). This means the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) results in \( \sin^{-1}(x) + C \), where \( C \) is the constant of integration.

Step 3: Apply the Constant Multiple Rule

Since there is a constant 5 multiplied by the integrand, we can factor the constant out of the integral. So we have: \( 5 \int \frac{1}{\sqrt{1-x^2}} \, dx \).

Step 4: Evaluate the Integral

Using the result from Step 2, \( \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C \), apply this to the integral: \( 5 \sin^{-1}(x) + C \).

Step 5: Write the Final Solution

The indefinite integral of \( \int \frac{5}{\sqrt{1-x^2}} \, dx \) is \( 5 \sin^{-1}(x) + C \), where \( C \) is the constant of integration.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential elements in calculus, as they help us find angles when given certain trigonometric values. This is particularly handy when working with integrals that resemble trigonometric functions. Inverse trigonometric functions have derivatives that you should memorize for solving integrals efficiently. For example:

• The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
• The derivative of \( \cos^{-1}(x) \) is \( -\frac{1}{\sqrt{1-x^2}} \).
• The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).

In our exercise, we recognize the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) as matching the derivative for \( \sin^{-1}(x) \). Thus, its antiderivative is \( \sin^{-1}(x) + C \). Recognizing these patterns allows us to transform complex integrals into simpler forms, making the process of integration much easier.

Integral Calculus

Integral calculus is a fundamental part of mathematics dealing with integrals. Unlike differential calculus, which focuses on the rate of change, integral calculus helps in finding total values like area under a curve. Indefinite integrals, like in our problem, lack specific limits to evaluate a particular numerical value.

• To solve indefinite integrals, you find the antiderivative or primitive of the function, expressing it as a function plus a constant of integration \( C \).
• Techniques often used include substitution, integration by parts, and recognizing standard forms, such as those involving inverse trigonometric functions.

In our exercise example, by identifying the integrand as a known derivative, we made the integration process straightforward. Recognizing this common form was the key to solving it. Mastery of integral calculus not only involves mechanical calculations but also an awareness of how derivatives and integrals are intertwined.

Constant of Integration

When dealing with indefinite integrals, the constant of integration \( C \) appears as an essential element. Why is this constant so crucial? Whenever you find the antiderivative of a function, many potential solutions exist, differing by a constant amount.

• The constant \( C \) acts as a placeholder for all possible these values.
• It ensures that every function that shares the same derivative is accounted for in the solution.
• This becomes vital when integrating functions because the original antiderivative could have shifted vertically by any constant.

In the exercise, when we arrive at the solution \( 5 \sin^{-1}(x) + C \), the \( C \) ensures we do not miss any potential solutions. Remember, in definite integrals, this constant cancels out when evaluating limits, but for indefinite integrals, it remains a fundamental aspect of the general solution.