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 $x$, $\lvert X\rvert < 1$.

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:

1. Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$.
2. 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 $2A(x)$.
3. 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:

1. Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$.
2. 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 $2A(x)$.
3. 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:

1. Show that the derivative of $A\left(\frac{2x}{1-x^2}\right)$ is equal to the derivative of $2A(x)$.
2. 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 $2A(x)$.
3. 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.