Why the Answer Varies Depending on u

Oct 13, 2025 by leolavanderia.com.br 14 views

In u-substitution method, why does the answer vary depending on what you choose as u?

The u-substitution method is a powerful technique used in calculus to solve definite and indefinite integrals. It involves substituting a new variable, u, into the original function to simplify the integration process. However, the choice of u can significantly impact the final result, and it's not uncommon for different u-substitutions to yield different answers. In this article, we'll explore why the answer varies depending on what you choose as u, using the integral $\int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx$ as a case study.

The given integral is $\int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx$ . At first glance, this integral appears to be difficult to solve using traditional integration techniques. However, with the u-substitution method, we can simplify the integral and arrive at a solution.

Method 1: Let $u=1-a^{2x}$

One possible u-substitution is to let $u=1-a^{2x}$ . This substitution is motivated by the fact that the denominator of the integral contains a square root, and we can simplify the expression by eliminating the square root.

To find $du$ , we differentiate $u$ with respect to $x$ :

\frac{du}{dx} = -2a^2 \ln a

We can rewrite this as:

du = -2a^2 \ln a \ dx

Now, we can substitute $u$ and $du$ into the original integral:

\int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx = \int\frac{a^x}{\sqrt{u}}\ \frac{1}{-2a^2 \ln a}\ du

Simplifying the expression, we get:

\int\frac{a^x}{\sqrt{u}}\ \frac{1}{-2a^2 \ln a}\ du = -\frac{1}{2\ln a}\int\frac{a^x}{\sqrt{u}}\ du

Now, we can evaluate the integral with respect to $u$ :

-\frac{1}{2\ln a}\int\frac{a^x}{\sqrt{u}}\ du = -\frac{1}{2\ln a}\left(2\sqrt{u}\right) + C

Substituting back $u=1-a^{2x}$ , we get:

-\frac{1}{2\ln a}\left(2\sqrt{u}\right) + C = -\frac{1}{\ln a}\sqrt{1-a^{2x}} + C

Method 2: Let $u=a^x$

Another possible u-substitution is to let $u=a^x$ . This substitution is motivated by the fact that the numerator of the integral contains a power of $a^x$ , and we can simplify the expression by eliminating the power.

To find $du$ , we differentiate $u$ with respect to $x$ :

\frac{du}{dx} = a^x \ln a

We can rewrite this as:

du = a^x \ln a \ dx

Now, we can substitute $u$ and $du$ into the original integral:

\int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx = \int\frac{u}{\sqrt{1-u^2}}\ \frac{1}{a^x \ln a}\ du

Simplifying the expression, we get:

\int\frac{u}{\sqrt{1-u^2}}\ \frac{1}{a^x \ln a}\ du = \frac{1}{\ln a}\int\frac{u}{\sqrt{1-u^2}}\ du

Now, we can evaluate the integral with respect to $u$ :

\frac{1}{\ln a}\int\frac{u}{\sqrt{1-u^2}}\ du = \frac{1}{\ln a}\left(\sqrt{1-u^2}\right) + C

Substituting back $u=a^x$ , we get:

\frac{1}{\ln a}\left(\sqrt{1-u^2}\right) + C = \frac{1}{\ln a}\sqrt{1-a^{2x}} + C

Why the Answer Varies Depending on u

As we can see from the two methods above, the answer to the integral varies depending on the choice of u. In Method 1, we let $u=1-a^{2x}$ , and in Method 2, we let $u=a^x$ . The resulting integrals are different, and the final answers are also different.

The reason for this variation lies in the fact that the u-substitution method is not a one-to-one correspondence between the original function and the new function. In other words, the same function can be represented in different ways using different u-substitutions.

When we choose a particular u-substitution, we are essentially changing the representation of the function. This change of representation can lead to different integrals and different final answers.

In conclusion, the u-substitution method is a powerful technique used in calculus to solve definite and indefinite integrals. However, the choice of u can significantly impact the final result, and it's not uncommon for different u-substitutions to yield different answers. By understanding the underlying reasons for this variation, we can better appreciate the nuances of the u-substitution method and use it more effectively in solving integrals.

The final answer to the integral $\int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx$ is:

-\frac{1}{\ln a}\sqrt{1-a^{2x}} + C

\frac{1}{\ln a}\sqrt{1-a^{2x}} + C

depending on the choice of u-substitution.
Q&A: Understanding the u-Substitution Method

The u-substitution method is a powerful technique used in calculus to solve definite and indefinite integrals. However, it can be challenging to understand and apply, especially when it comes to choosing the right u-substitution. In this article, we'll answer some frequently asked questions about the u-substitution method to help you better understand and apply it.

Q: What is the u-substitution method?

A: The u-substitution method is a technique used in calculus to solve definite and indefinite integrals by substituting a new variable, u, into the original function. This substitution can simplify the integration process and make it easier to solve the integral.

Q: Why do we need to choose a u-substitution?

A: We need to choose a u-substitution because the u-substitution method is not a one-to-one correspondence between the original function and the new function. In other words, the same function can be represented in different ways using different u-substitutions. The choice of u-substitution can significantly impact the final result, and it's not uncommon for different u-substitutions to yield different answers.

Q: How do I choose a u-substitution?

A: Choosing a u-substitution involves identifying a part of the original function that can be simplified or eliminated using a substitution. You can use various techniques, such as factoring, canceling, or rearranging terms, to identify a suitable u-substitution.

Q: What are some common u-substitutions?

A: Some common u-substitutions include:

Letting u be a function of x, such as u = f(x)
Letting u be a constant multiple of x, such as u = cx
Letting u be a power of x, such as u = x^n
Letting u be a product of x and another function, such as u = x*f(x)

Q: How do I find du?

A: To find du, you need to differentiate the u-substitution with respect to x. This will give you an expression for du in terms of x and the original function.

Q: What if I make a mistake in choosing a u-substitution?

A: If you make a mistake in choosing a u-substitution, you may end up with a more complicated integral or a different final answer. However, you can always try again with a different u-substitution or use a different technique to solve the integral.

Q: Can I use the u-substitution method for all types of integrals?

A: The u-substitution method is most effective for integrals that involve trigonometric functions, exponential functions, or rational functions. However, it can also be used for other types of integrals, such as integrals involving logarithmic functions or absolute value functions.

Q: How do I know if I've chosen the right u-substitution?

A: You can check if you've chosen the right u-substitution by substituting the u-substitution back into the original function and simplifying the expression. If the resulting expression is simpler and easier to integrate, then you've likely chosen the right u-substitution.

In conclusion, the u-substitution method is a powerful technique used in calculus to solve definite and indefinite integrals. By understanding the underlying principles and techniques of the u-substitution method, you can better appreciate its strengths and limitations and use it more effectively in solving integrals.

Choosing a u-substitution that is too complicated or difficult to work with
Failing to check if the u-substitution is a one-to-one correspondence between the original function and the new function
Not simplifying the expression after substituting the u-substitution back into the original function
Not checking if the resulting expression is simpler and easier to integrate

Use the u-substitution method when the integral involves trigonometric functions, exponential functions, or rational functions
Choose a u-substitution that simplifies the expression and makes it easier to integrate
Use the chain rule to find du when the u-substitution involves a composite function
Check if the u-substitution is a one-to-one correspondence between the original function and the new function
Simplify the expression after substituting the u-substitution back into the original function