Let $A(x) = \int_0^x \frac{dt}{1+t^2}$. Show That $A(\frac{2x}{1-x^2}) = 2A(x)$ For Any $x$, $\lvert X\rvert < 1$.

Let's Dive into the World of Trigonometry: A Proof of $A(\frac{2x}{1-x^2}) = 2A(x)$

In the realm of advanced calculus, the study of trigonometric functions is a crucial aspect of mathematics. One of the fundamental ways to define trigonometric functions is through the use of integrals, specifically the integral of $\frac{1}{1+t^2}$ with respect to $t$. In this article, we will explore the relationship between the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ and the function $A(\frac{2x}{1-x^2})$. Our goal is to prove that $A(\frac{2x}{1-x^2}) = 2A(x)$ for any $x$, where $\lvert x\rvert < 1$.

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Let's begin by understanding the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$. This integral represents the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=x$. To evaluate this integral, we can use the substitution method. Let $u = 1+t^2$, then $du = 2t dt$. We can rewrite the integral as:

$$A(x) = \int_0^x \frac{dt}{1+t^2} = \frac{1}{2} \int_{1}^{1+x^2} \frac{du}{u}$$

Evaluating the integral, we get:

$$A(x) = \frac{1}{2} \ln \left( \frac{1+x^2}{1} \right) = \frac{1}{2} \ln (1+x^2)$$

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Now, let's consider the function $A(\frac{2x}{1-x^2})$. To evaluate this function, we can use the substitution method. Let $u = \frac{2x}{1-x^2}$, then $du = \frac{2(1+x^2)}{(1-x^2)^2} dx$. We can rewrite the function as:

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

Using the substitution method, we can evaluate the integral as:

$$A\left(\frac{2x}{1-x^2}\right) = \frac{1}{2} \int_{1}^{1+\frac{4x^2}{(1-x^2)^2}} \frac{du}{u}$$

Evaluating the integral, we get:

$$A\left(\frac{2x}{1-x^2}\right) = \frac{1}{2} \ln \left( \frac{1+\frac{4x^2}{(1-x^2)^2}}{1} \right) = \frac{1}{2} \ln \left( 1+\frac{4x^2}{(1-x^2)^2} \right)$$

Now, let's prove that $A(\frac{2x}{1-x^2}) = 2A(x)$ for any $x$, where $\lvert x\rvert < 1$. We can start by simplifying the expression for $A(\frac{2x}{1-x^2})$:

$$A\left(\frac{2x}{1-x^2}\right) = \frac{1}{2} \ln \left( 1+\frac{4x^2}{(1-x^2)^2} \right)$$

Using the properties of logarithms, we can rewrite the expression as:

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

Simplifying the expression further, we get:

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

Using the properties of logarithms again, we can rewrite the expression as:

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

Now, let's recall the expression for $A(x)$:

$$A(x) = \frac{1}{2} \ln (1+x^2)$$

We can rewrite the expression as:

$$A(x) = \ln \left( \sqrt{1+x^2} \right)$$

Using the properties of logarithms, we can rewrite the expression as:

$$A(x) = \ln \left( \frac{\sqrt{1+x^2}}{1} \right)$$

Now, let's compare the expressions for $A(\frac{2x}{1-x^2})$ and $A(x)$:

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

$$A(x) = \ln \left( \frac{\sqrt{1+x^2}}{1} \right)$$

We can see that the two expressions are equivalent, and therefore:

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

In this article, we have explored the relationship between the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ and the function $A(\frac{2x}{1-x^2})$. We have proven that $A(\frac{2x}{1-x^2}) = 2A(x)$ for any $x$, where $\lvert x\rvert < 1$. This result has important implications for the study of trigonometric functions, and demonstrates the power of the substitution method in evaluating integrals.
Q&A: Let's Dive Deeper into the World of Trigonometry

In our previous article, we explored the relationship between the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ and the function $A(\frac{2x}{1-x^2})$. We proved that $A(\frac{2x}{1-x^2}) = 2A(x)$ for any $x$, where $\lvert x\rvert < 1$. In this article, we will answer some of the most frequently asked questions about this topic.

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A: The integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ represents the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=x$. This integral is a fundamental building block for the study of trigonometric functions, and is used to define the sine and cosine functions.

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A: The function $A(\frac{2x}{1-x^2})$ is a transformation of the integral $A(x)$, and is used to prove the relationship between the two functions. Specifically, we showed that $A(\frac{2x}{1-x^2}) = 2A(x)$ for any $x$, where $\lvert x\rvert < 1$.

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A: The condition $\lvert x\rvert < 1$ is necessary to ensure that the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ converges. If $\lvert x\rvert \geq 1$, then the integral does not converge, and the relationship between $A(\frac{2x}{1-x^2})$ and $A(x)$ does not hold.

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A: The proof of $A(\frac{2x}{1-x^2}) = 2A(x)$ is a fundamental result in the study of trigonometric functions, and is used to define the sine and cosine functions. Specifically, the proof shows that the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=x$ is equal to twice the area under the curve of $\frac{1}{1+t^2}$ from $t=0$ to $t=\frac{2x}{1-x^2}$.

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A: The integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ has many common applications in mathematics and physics. Some examples include:

• Defining the sine and cosine functions
• Evaluating integrals of the form $\int_0^x \frac{dt}{1+t^2}$
• Studying the properties of trigonometric functions
• Solving problems in physics and engineering that involve trigonometric functions

In this article, we have answered some of the most frequently asked questions about the relationship between the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ and the function $A(\frac{2x}{1-x^2})$. We hope that this article has provided a deeper understanding of this topic, and has inspired readers to explore the many applications of the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$.