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

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 requires us to show that a given function $A(x)$, defined as the integral of $\frac{1}{1+t^2}$ from $0$ to $x$, can be expressed as $2A(x)$ when the variable is replaced by $\frac{2x}{1-x^2}$. We will use the technique of change of variable to solve this problem.

To solve this problem, we will use the technique of change of variable. This technique involves substituting a new variable into the integral, which can simplify the integral and make it easier to evaluate. In this case, we will substitute $t = \tan\theta$ into the integral.

Substitution

Let $t = \tan\theta$. Then, we have $dt = \sec^2\theta d\theta$. We also have $1 + t^2 = 1 + \tan^2\theta = \sec^2\theta$.

Integral

The integral becomes:

$$\int_0^x \frac{dt}{1+t^2} = \int_0^{\tan^{-1}x} \frac{\sec^2\theta d\theta}{\sec^2\theta} = \int_0^{\tan^{-1}x} d\theta = \tan^{-1}x$$

Expression for $A(x)$

We can now express $A(x)$ as:

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

Change of Variable

Now, we will substitute $t = \tan\theta$ into the expression for $A\left(\frac{2x}{1-x^2}\right)$.

Substitution

Let $t = \tan\theta$. Then, we have $dt = \sec^2\theta d\theta$. We also have $1 + t^2 = 1 + \tan^2\theta = \sec^2\theta$.

Integral

The integral becomes:

$$\int_0^{\frac{2x}{1-x^2}} \frac{dt}{1+t^2} = \int_0^{\tan^{-1}\left(\frac{2x}{1-x^2}\right)} \frac{\sec^2\theta d\theta}{\sec^2\theta} = \int_0^{\tan^{-1}\left(\frac{2x}{1-x^2}\right)} d\theta = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$$

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

We can now express $A\left(\frac{2x}{1-x^2}\right)$ as:

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

Relationship between $A(x)$ and $A\left(\frac{2x}{1-x^2}\right)$

We can now establish a relationship between $A(x)$ and $A\left(\frac{2x}{1-x^2}\right)$.

Derivation

We have:

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

We can also express $\frac{2x}{1-x^2}$ as:

$$\frac{2x}{1-x^2} = \tan(2\theta)$$

where $\theta = \tan^{-1}x$.

Substitution

Let $\theta = \tan^{-1}x$. Then, we have:

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

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

We can now express $A\left(\frac{2x}{1-x^2}\right)$ as:

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

Conclusion

We have shown that $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ for any $x$ such that $\lvert x\rvert < 1$. This result can be used to simplify the evaluation of definite integrals involving the function $\frac{1}{1+t^2}$.

The final answer is: $\boxed{2A(x)}$
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 $x$, $\lvert x\rvert < 1$

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A: The function $A(x)$ is defined as the integral of $\frac{1}{1+t^2}$ from $0$ to $x$. In other words, $A(x) = \int_0^x \frac{dt}{1+t^2}$.

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A: We have shown that $A\left(\frac{2x}{1-x^2}\right) = 2A(x)$ for any $x$ such that $\lvert x\rvert < 1$.

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A: We used the technique of change of variable to derive the relationship between $A(x)$ and $A\left(\frac{2x}{1-x^2}\right)$. We substituted $t = \tan\theta$ into the integral and then used the trigonometric identity $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$ to simplify the expression.

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A: The condition $\lvert x\rvert < 1$ is necessary to ensure that the function $A(x)$ is well-defined. If $\lvert x\rvert \geq 1$, then the integral $\int_0^x \frac{dt}{1+t^2}$ may not converge.

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A: Yes, here is an example. Suppose we want to evaluate the definite integral $\int_0^1 \frac{dt}{1+t^2}$. We can use the relationship between $A(x)$ and $A\left(\frac{2x}{1-x^2}\right)$ to simplify the evaluation of this integral.

Substitution

Let $x = \frac{2t}{1+t^2}$. Then, we have:

$$\int_0^1 \frac{dt}{1+t^2} = \int_0^1 \frac{1+t^2}{1+t^2} dt = \int_0^1 1 dt = 1$$

Expression for $A(1)$

We can now express $A(1)$ as:

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

Relationship between $A(1)$ and $A\left(\frac{2}{1-1}\right)$

We can now establish a relationship between $A(1)$ and $A\left(\frac{2}{1-1}\right)$.

Derivation

We have:

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

We can also express $\frac{2}{1-1}$ as:

$$\frac{2}{1-1} = \tan(\infty)$$

where $\theta = \tan^{-1}1$.

Substitution

Let $\theta = \tan^{-1}1$. Then, we have:

$$A\left(\frac{2}{1-1}\right) = \tan^{-1}\left(\tan(\infty)\right) = \frac{\pi}{2}$$

Expression for $A\left(\frac{2}{1-1}\right)$

We can now express $A\left(\frac{2}{1-1}\right)$ as:

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

Relationship between $A(1)$ and $A\left(\frac{2}{1-1}\right)$

We can now establish a relationship between $A(1)$ and $A\left(\frac{2}{1-1}\right)$.

Derivation

We have:

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

We also have:

$$A(1) = 1$$

Conclusion

We have shown that $A\left(\frac{2}{1-1}\right) = \frac{\pi}{2}$ and $A(1) = 1$. This result can be used to simplify the evaluation of definite integrals involving the function $\frac{1}{1+t^2}$.

The final answer is: $\boxed{\frac{\pi}{2}}$