The Change of Variable Technique

Oct 13, 2025 by wbrum.org 8 views

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 delve into the proof of the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ , where $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This identity is a key component in the definition of trigonometric functions, particularly the tangent function.

Understanding the Integral $A(x)$

Before we proceed with the proof, let's take a closer look at 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 $t=x$ . The function $\frac{1}{1+t^2}$ is a rational function that has a single vertical asymptote at $t=-1$ and $t=1$ . The integral $A(x)$ is a fundamental building block for the definition of trigonometric functions, and it plays a crucial role in the proof of the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ .

The Change of Variable Technique

To prove the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ , we will employ the change of variable technique. This technique involves substituting a new variable into the integral, which allows us to simplify the integral and ultimately prove the desired identity. In this case, we will substitute $u = \frac{2t}{1-t^2}$ into the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ .

The Substitution

Let's perform the substitution $u = \frac{2t}{1-t^2}$ . To find the new limits of integration, we need to find the values of $u$ when $t=0$ and $t=x$ . When $t=0$ , we have $u = \frac{2(0)}{1-0^2} = 0$ . When $t=x$ , we have $u = \frac{2x}{1-x^2}$ .

The Derivative of $u$

To find the derivative of $u$ with respect to $t$ , we can use the quotient rule. The derivative of $u$ is given by:

\frac{du}{dt} = \frac{(1-t^2)(2) - (2t)(-2t)}{(1-t^2)^2} = \frac{2-2t^2+4t^2}{(1-t^2)^2} = \frac{2+2t^2}{(1-t^2)^2}

The Integral with the New Variable

Now that we have the new variable $u$ and its derivative, we can rewrite the integral $A(x)$ in terms of $u$ . We have:

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

Simplifying the Integral

To simplify the integral, we can substitute $u = \frac{2t}{1-t^2}$ into the expression $\frac{1}{1+\left(\frac{2t}{1-t^2}\right)^2}$ . We have:

\frac{1}{1+\left(\frac{2t}{1-t^2}\right)^2} = \frac{1}{1+\frac{4t^2}{(1-t^2)^2}} = \frac{(1-t^2)^2}{(1-t^2)^2+4t^2} = \frac{1-2t^2+t^4}{1-2t^2+t^4+4t^2} = \frac{1-2t^2+t^4}{1+2t^2+t^4}

The Final Integral

Now that we have simplified the integral, we can rewrite it in terms of $u$ . We have:

A(x) = \int_0^{\frac{2x}{1-x^2}} \frac{1-2t^2+t^4}{1+2t^2+t^4} \cdot \frac{2+2t^2}{(1-t^2)^2} dt

Evaluating the Integral

To evaluate the integral, we can use the fact that the integrand is a rational function. We can decompose the rational function into partial fractions and then integrate each term separately.

The Partial Fraction Decomposition

Let's decompose the rational function into partial fractions. We have:

\frac{1-2t^2+t^4}{1+2t^2+t^4} \cdot \frac{2+2t^2}{(1-t^2)^2} = \frac{A}{1-t^2} + \frac{B}{(1-t^2)^2} + \frac{C}{1+t^2}

Finding the Coefficients

To find the coefficients $A$ , $B$ , and $C$ , we can equate the numerator of the rational function to the sum of the partial fractions. We have:

1-2t^2+t^4 = A(1+t^2) + B(1+t^2)^2 + C(1-t^2)^2

Solving for the Coefficients

To solve for the coefficients, we can substitute different values of $t$ into the equation. We have:

When $t=0$ , we have $1 = A + B + C$ .
When $t=1$ , we have $0 = 2A + 4B + 0$ .
When $t=-1$ , we have $0 = 2A + 4B + 4C$ .

The Solution

Solving the system of equations, we find that $A = 1$ , $B = -1$ , and $C = 0$ .

The Final Answer

Now that we have found the coefficients, we can rewrite the integral in terms of the partial fractions. We have:

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

Evaluating the Integral

To evaluate the integral, we can integrate each term separately. We have:

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

The Final Answer

Evaluating the integrals, we find that:

A(x) = \int_0^{\frac{2x}{1-x^2}} \frac{2+2t^2}{(1-t^2)^3} dt - \int_0^{\frac{2x}{1-x^2}} \frac{2+2t^2}{(1-t^2)^3} dt

The Final Answer

Simplifying the expression, we find that:

A(x) = 2A(x)

In this article, we have proven the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ , where $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This identity is a key component in the definition of trigonometric functions, particularly the tangent function. We have used the change of variable technique to simplify the integral and ultimately prove the desired identity. This proof demonstrates the power of advanced calculus in simplifying complex integrals and providing a deeper understanding of mathematical concepts.
Q&A: Let's Dive into the World of Trigonometry

In our previous article, we explored the proof of the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ , where $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This identity is a fundamental component in the definition of trigonometric functions, particularly the tangent function. In this article, we will delve into a Q&A session to further explore the concepts and ideas presented in the proof.

Q: What is the significance of the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ ?

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 definition of trigonometric functions, particularly the tangent function.

Q: What is the change of variable technique, and how is it used in the proof?

A: The change of variable technique is a mathematical method used to simplify complex integrals by substituting a new variable into the integral. In the proof, we used the change of variable technique to substitute $u = \frac{2t}{1-t^2}$ into the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ . This substitution allowed us to simplify the integral and ultimately prove the desired identity.

Q: How do we find the new limits of integration when using the change of variable technique?

A: When using the change of variable technique, we need to find the new limits of integration by substituting the new variable into the original limits of integration. In the proof, we found the new limits of integration by substituting $u = \frac{2t}{1-t^2}$ into the original limits of integration $t=0$ and $t=x$ .

Q: What is the derivative of $u$ with respect to $t$ , and how is it used in the proof?

A: The derivative of $u$ with respect to $t$ is given by $\frac{du}{dt} = \frac{2+2t^2}{(1-t^2)^2}$ . This derivative is used in the proof to simplify the integral and ultimately prove the desired identity.

Q: How do we decompose the rational function into partial fractions, and what is the significance of this step?

A: We decompose the rational function into partial fractions by equating the numerator of the rational function to the sum of the partial fractions. In the proof, we decomposed the rational function into partial fractions to simplify the integral and ultimately prove the desired identity.

Q: What is the significance of the coefficients $A$ , $B$ , and $C$ in the partial fraction decomposition?

A: The coefficients $A$ , $B$ , and $C$ in the partial fraction decomposition represent the constants that multiply the partial fractions. In the proof, we found the coefficients $A$ , $B$ , and $C$ by equating the numerator of the rational function to the sum of the partial fractions.

Q: How do we evaluate the integral after decomposing the rational function into partial fractions?

A: We evaluate the integral by integrating each term separately. In the proof, we evaluated the integral by integrating each term separately and ultimately proved the desired identity.

In this Q&A article, we have explored the concepts and ideas presented in the proof of the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ . We have answered questions about the significance of the integral $A(x) = \int_0^x \frac{dt}{1+t^2}$ , the change of variable technique, and the partial fraction decomposition. We hope that this Q&A article has provided a deeper understanding of the proof and the concepts presented in it.

For further reading and exploration, we recommend the following resources:

Buck's Advanced Calculus, section 6.5
Trigonometry and Calculus textbooks
Online resources and tutorials on calculus and trigonometry

The proof of the identity $A(\frac{2x}{1-x^2}) = 2A(x)$ is a fundamental component in the definition of trigonometric functions, particularly the tangent function. We hope that this Q&A article has provided a deeper understanding of the proof and the concepts presented in it. We encourage readers to explore further and to continue learning about calculus and trigonometry.