Simplifying the Transformed Function

Oct 13, 2025 by recoveryshake.com 126 views

Let $A(x) = \int_0^x \frac{dt}{1+t^2}$ . Show that $A\left(\frac{2x}{1-x^2}\right)= 2A(x)$ for any $x$ , $\lvert x\rvert < 1$

Introduction

In this article, we will explore a problem from Buck's Advanced Calculus, section 6.5, which involves a change of variable in a definite integral. The problem asks us to show that $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ for any $x$ such that $\lvert x\rvert < 1$ , where $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This problem is a great example of how a change of variable can be used to simplify a definite integral and reveal a deeper connection between the original function and the transformed function.

The Original Function

Let's start by examining the original function $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This function represents the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=x$ . The integrand $\frac{1}{1+t^2}$ is a well-known function that has a simple antiderivative, which is $\arctan(t)$ . Therefore, we can write the original function as:

A(x) = \int_0^x \frac{dt}{1+t^2} = \arctan(x) - \arctan(0) = \arctan(x)

The Transformed Function

Now, let's examine the transformed function $A\left(\frac{2x}{1-x^2}\right)$ . This function represents the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=\frac{2x}{1-x^2}$ . To simplify this expression, we can use a change of variable. Let $u = \frac{2x}{1-x^2}$ . Then, we can write:

A\left(\frac{2x}{1-x^2}\right) = \int_0^{\frac{2x}{1-x^2}} \frac{dt}{1+t^2} = \int_0^u \frac{du}{1+u^2}

Change of Variable

To evaluate the integral $\int_0^u \frac{du}{1+u^2}$ , we can use a change of variable. Let $v = \arctan(u)$ . Then, we can write:

\int_0^u \frac{du}{1+u^2} = \int_0^{\arctan(u)} \frac{dv}{1+\tan^2(v)}

Using the trigonometric identity $\tan^2(v) + 1 = \sec^2(v)$ , we can simplify the integral to:

\int_0^{\arctan(u)} \frac{dv}{1+\tan^2(v)} = \int_0^{\arctan(u)} \frac{dv}{\sec^2(v)} = \arctan(u) - \arctan(0) = \arctan(u)

Simplifying the Transformed Function

Now, let's simplify the transformed function $A\left(\frac{2x}{1-x^2}\right)$ . We can write:

A\left(\frac{2x}{1-x^2}\right) = \int_0^u \frac{du}{1+u^2} = \arctan(u)

Using the definition of $u$ , we can write:

A\left(\frac{2x}{1-x^2}\right) = \arctan\left(\frac{2x}{1-x^2}\right)

Connection to the Original Function

Now, let's examine the connection between the original function $A(x) = \arctan(x)$ and the transformed function $A\left(\frac{2x}{1-x^2}\right) = \arctan\left(\frac{2x}{1-x^2}\right)$ . We can see that the transformed function is a composition of the original function with the function $\frac{2x}{1-x^2}$ . This suggests that the transformed function is a scaling and shifting of the original function.

Proof of the Identity

To prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ , we can use the following steps:

Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$ .
Use the fact that the derivative of an integral is equal to the integrand to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the integral of the derivative of $2A(x)$ .
Use the fundamental theorem of calculus to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to $A\left(\frac{2x}{1-x^2}\right)$ and the integral of the derivative of $2A(x)$ is equal to $2A(x)$ .

Derivative of the Transformed Function

To find the derivative of the transformed function $A\left(\frac{2x}{1-x^2}\right)$ , we can use the chain rule:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{d}{du} \arctan(u) \cdot \frac{du}{dx}

Using the definition of $u$ , we can write:

\frac{du}{dx} = \frac{2}{1-x^2} - \frac{2x(2)}{(1-x^2)^2} = \frac{2(1-x^2) - 2x(2)}{(1-x^2)^2} = \frac{2-2x^2-4x}{(1-x^2)^2}

Simplifying the expression, we get:

\frac{du}{dx} = \frac{2(1-2x-x^2)}{(1-x^2)^2}

Derivative of the Original Function

To find the derivative of the original function $2A(x)$ , we can use the chain rule:

\frac{d}{dx} 2A(x) = 2 \cdot \frac{d}{dx} \arctan(x)

Using the fact that the derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$ , we get:

\frac{d}{dx} 2A(x) = 2 \cdot \frac{1}{1+x^2}

Proof of the Identity

Now, let's prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ . We can start by showing that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$ .

Using the chain rule, we can write:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{d}{du} \arctan(u) \cdot \frac{du}{dx} = \frac{1}{1+u^2} \cdot \frac{2(1-2x-x^2)}{(1-x^2)^2}

Using the definition of $u$ , we can write:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{1}{1+\left(\frac{2x}{1-x^2}\right)^2} \cdot \frac{2(1-2x-x^2)}{(1-x^2)^2}

Simplifying the expression, we get:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{1}{1+\frac{4x^2}{(1-x^2)^2}} \cdot \frac{2(1-2x-x^2)}{(1-x^2)^2}

Using the fact that $1+\frac{4x^2}{(1-x^2)^2} = \frac{(1-x^2)^2+4x^2}{(1-x^2)^2} = \frac{1-2x^2+4x^2}{(1-x^2)^2} = \frac{1+2x^2}{(1-x^2)^2}$ , we can simplify the expression to:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{1}{\frac{1+2x^2}{(1-x^2)^2}} \cdot \frac{2(1-2x-x^2)}{(1-x^2)^2} = \frac{(1-x^2)^2<br/> **Q&A: Let $A(x) = \int_0^x \frac{dt}{1+t^2}$. Show that $A\left(\frac{2x}{1-x^2}\right)= 2A(x)$ for any $x$, $\lvert x\rvert < 1$** ### Q: What is the original function $A(x)$? A: The original function $A(x)$ is defined as the integral of $\frac{1}{1+t^2}$ from $t=0$ to $t=x$. This can be written as: $$A(x) = \int_0^x \frac{dt}{1+t^2}

Q: What is the transformed function $A\left(\frac{2x}{1-x^2}\right)$ ?

A: The transformed function $A\left(\frac{2x}{1-x^2}\right)$ is defined as the integral of $\frac{1}{1+t^2}$ from $t=0$ to $t=\frac{2x}{1-x^2}$ . This can be written as:

A\left(\frac{2x}{1-x^2}\right) = \int_0^{\frac{2x}{1-x^2}} \frac{dt}{1+t^2}

Q: How do we simplify the transformed function?

A: To simplify the transformed function, we can use a change of variable. Let $u = \frac{2x}{1-x^2}$ . Then, we can write:

A\left(\frac{2x}{1-x^2}\right) = \int_0^u \frac{du}{1+u^2}

Q: How do we evaluate the integral of the transformed function?

A: To evaluate the integral of the transformed function, we can use the fact that the derivative of $\arctan(u)$ is $\frac{1}{1+u^2}$ . Therefore, we can write:

\int_0^u \frac{du}{1+u^2} = \arctan(u) - \arctan(0) = \arctan(u)

Q: What is the connection between the original function and the transformed function?

A: The transformed function is a composition of the original function with the function $\frac{2x}{1-x^2}$ . This suggests that the transformed function is a scaling and shifting of the original function.

Q: How do we prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ ?

A: To prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ , we can use the following steps:

Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$ .
Use the fact that the derivative of an integral is equal to the integrand to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the integral of the derivative of $2A(x)$ .
Use the fundamental theorem of calculus to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to $A\left(\frac{2x}{1-x^2}\right)$ and the integral of the derivative of $2A(x)$ is equal to $2A(x)$ .

Q: What is the derivative of the transformed function?

A: The derivative of the transformed function $A\left(\frac{2x}{1-x^2}\right)$ is given by:

\frac{d}{dx} A\left(\frac{2x}{1-x^2}\right) = \frac{d}{du} \arctan(u) \cdot \frac{du}{dx}

Using the definition of $u$ , we can write:

\frac{du}{dx} = \frac{2}{1-x^2} - \frac{2x(2)}{(1-x^2)^2} = \frac{2(1-x^2) - 2x(2)}{(1-x^2)^2} = \frac{2-2x^2-4x}{(1-x^2)^2}

Simplifying the expression, we get:

\frac{du}{dx} = \frac{2(1-2x-x^2)}{(1-x^2)^2}

Q: What is the derivative of the original function?

A: The derivative of the original function $2A(x)$ is given by:

\frac{d}{dx} 2A(x) = 2 \cdot \frac{d}{dx} \arctan(x)

Using the fact that the derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$ , we get:

\frac{d}{dx} 2A(x) = 2 \cdot \frac{1}{1+x^2}

Q: How do we prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ ?

A: To prove the identity $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ , we can use the following steps:

Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$ .
Use the fact that the derivative of an integral is equal to the integrand to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the integral of the derivative of $2A(x)$ .
Use the fundamental theorem of calculus to show that the integral of the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to $A\left(\frac{2x}{1-x^2}\right)$ and the integral of the derivative of $2A(x)$ is equal to $2A(x)$ .

Q: What is the final answer?

A: The final answer is:

A\left(\frac{2x}{1-x^2}\right) = 2A(x)

This identity shows that the transformed function $A\left(\frac{2x}{1-x^2}\right)$ is equal to twice the original function $A(x)$ . This is a useful result that can be used to simplify and evaluate definite integrals.