# Integral calculus: integral variable substitution confusion

• thegreengineer
In summary, the conversation discusses the process of solving integrals using variable substitution, specifically in cases involving fractional powers and trigonometric functions. There is a question regarding the method used by the math teacher in finding the value of du.
thegreengineer
Recently I started seeing integral calculus and right now we are covering the topic of the antiderivative. At first sign it was not very difficult, until we started seeing integral variable substitution. The problem starts right here:
Let's suppose that we have a function like this:

$\int \sqrt{x+2}dx$

We can easily convert that $\sqrt{x+2}$ into a $(x+2)^{1/2}$. However we cannot expand this polynomial to a fractional power so instead we rename $x+2$ as $u$ so $x+2=u$ and $dx$ would become into $du$. Having this we can redefine the integral to have something as this:

$\int u^{1/2}du$

Now we can use the power rule to solve this integral:

$\frac{2}{3}u^{3/2}+C$

The only thing we now have to do is to express the solution in terms of $x$ so we have that:

$\frac{2}{3}(x+2)^{3/2}+C$

And this was just the beginning; later on we saw more difficult examples involving trigonometric functions. My main doubt focuses on what my math teacher did in class.

We started having this another integral:

$\int 2x\sqrt{4x-3}dx$

It's a very similar problem like the previous one, now having a factor multiplying a square root. Following the similar algorithm for the last case I replaced the whole expression $4x-3$ into $u$ and to express the remaining $x$ in terms of $u$ I only isolated for $x$ in the $u$. However what my teacher did to find $du$ was equalling it to $4dx$ and that's where I don't understand why instead of $du=dx$ he did that. I hope someone can answer me this to solve my homework problems. Thanks

The substitution was ##u = 4x - 3## and hence ##\frac{du}{dx} = 4##. Your teacher is then rearranging to get ##du = 4 \, dx##
(I personally don't like this method, the derivative is not a fraction of du divided by dx).

thegreengineer
pwsnafu said:
The substitution was ##u = 4x - 3## and hence ##\frac{du}{dx} = 4##. Your teacher is then rearranging to get ##du = 4 \, dx##
(I personally don't like this method, the derivative is not a fraction of du divided by dx).
I agree that technically du/dx is not a fraction, but most of the time it does no harm to treat it as if it were. This sort of arithmetic is done all the time in integration substitutions, including trig substitutions, as well as in integration by parts.

## 1. What is integral variable substitution in calculus?

Integral variable substitution is a technique used in integral calculus to simplify the integration of complex functions. It involves replacing the variable in the function with a new variable in order to make the integral easier to solve.

## 2. How do I know when to use integral variable substitution?

Integral variable substitution is typically used when the integrand (the function being integrated) contains a composition of functions or a product of functions. It can also be used when the integrand contains a radical expression.

## 3. What are some common mistakes made when using integral variable substitution?

Some common mistakes include forgetting to change the limits of integration, choosing an inappropriate substitution, and not simplifying the integrand after substitution. It is important to carefully follow the steps of integral variable substitution to avoid these errors.

## 4. Can I use any variable for substitution in integral calculus?

Yes, you can use any variable for substitution in integral calculus. However, it is recommended to choose a variable that simplifies the integrand and makes the integration process easier.

## 5. Are there any tips for mastering integral variable substitution?

Practice and familiarize yourself with different types of integrals and their corresponding substitutions. It is also helpful to check your answer by differentiating it to ensure that it is correct. With practice, integral variable substitution will become easier and more intuitive.

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