Find the Antiderivative f(x)=12/( square root of 1-x^2)

Solution:

Step 1:

Identify the antiderivative $F(x)$ by integrating the given function $f(x)$.

$$F(x)=\int f(x)dx$$

Step 2:

Write down the integral that needs to be solved.

$$F(x)=\int \frac{12}{\sqrt{1-x^2}}dx$$

Step 3:

Extract the constant $12$ from the integral as it does not depend on $x$.

$$12\int \frac{1}{\sqrt{1-x^2}}dx$$

Step 4:

Express the number $1$ as a square, $1^2$, to match the form of a trigonometric identity.

$$12\int \frac{1}{\sqrt{1^2-x^2}}dx$$

Step 5:

Recognize the integral as the inverse sine function, $\arcsin (x)$.

$$12(\arcsin (x)+C)$$

Step 6:

Combine the constant multiple with the antiderivative.

$$12\arcsin (x)+C$$

Step 7:

Conclude with the antiderivative of the function $f(x)=\frac{12}{\sqrt{1-x^2}}$.

$$F(x)=12\arcsin (x)+C$$

Knowledge Notes:

The process of finding the antiderivative involves integrating the given function. The integral of a function $f(x)$ is denoted by $\int f(x)dx$, and it represents the accumulation of the area under the curve of $f(x)$.

In this particular problem, we are dealing with the integral of a function that resembles the derivative of the arcsine function. The general form of the integral that corresponds to the arcsine function is $\int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+C$, where $C$ is the constant of integration.

The constant multiple rule in integration states that if $k$ is a constant and $f(x)$ is an integrable function, then $\int k\cdot f(x)dx=k\cdot \int f(x)dx$. This rule allows us to move constants outside the integral.

The antiderivative found through this process is not unique; it includes an arbitrary constant $C$, because the derivative of a constant is zero. This constant represents the family of all antiderivatives of the function.

The inverse trigonometric function $\arcsin (x)$ is the antiderivative of $\frac{1}{\sqrt{1-x^2}}$ within the domain $-1\le x\le 1$. It is important to recognize this integral form to solve the problem efficiently.