# Integration by parts question

• mrg
In summary, the conversation is about finding the integral of x^3*cos(x^2) and the attempt at solving it using integration by parts. The person tried a different method first but ended up with an incorrect answer. They asked if it matters what variables are chosen for u and dv, to which the other person replied that it does matter and pointed out an error in the work. The conversation ends with the person thanking the other for their help.

#### mrg

Homework Statement
$$\int x^{3}cos(x^{2})dx$$

The attempt at a solution
OK, so I am aware that there is a way in which to do this problem where you do a substitution (let $$u=x^{2}$$ to do a substitution before you integrate by parts), and I was able to get the answer right using this method. The thing is, I tried it a different way first, and after triple checking it, I feel like I should have gotten it right. It doesn't matter what I choose to let u and dv equal, right? I should get the right answer no matter what I choose (assuming I did things correctly, of course).
Here's my work:

$$u=cos(x^{2})$$
$$du=-2xsin(x^{2})$$
$$v=\frac{x^{4}}{4}$$
$$dv=x^{3}dx$$

$$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}\int x*sin(x^{2})dx$$

$$u_2=sin(x^{2})$$
$$du_2=2xcos(x)$$
$$v_2=\frac{x^{2}}{2}$$
$$dv_2=x$$

$$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}(\frac{x^{2}sin(x^{2})}{2}-\int x^{3}cos(x^{2})dx)$$

Since this last integral is the same as the one we started with, we can now say that
$$I=A+B-\frac{1}{2}I$$
Hence
$$\frac{3}{2}I=A+B$$
Multiplying both sides by two over three...
$$\frac{x^{4}cos(x^{2})}{6}+\frac{x^{2}sin(x^{2})}{6}$$Thanks for your time and any help you can offer me.

mrg said:
Homework Statement
$$\int x^{3}cos(x^{2})dx$$
The attempt at a solution
OK, so I am aware that there is a way in which to do this problem where you do a substitution (let $$u=x^{2}$$ to do a substitution before you integrate by parts), and I was able to get the answer right using this method. The thing is, I tried it a different way first, and after triple checking it, I feel like I should have gotten it right. It doesn't matter what I choose to let u and dv equal, right? I should get the right answer no matter what I choose (assuming I did things correctly, of course).
Here's my work:
$$u=cos(x^{2})$$
$$du=-2xsin(x^{2})$$
$$v=\frac{x^{4}}{4}$$
$$dv=x^{3}dx$$

$$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}\int x*sin(x^{2})dx$$...

##\displaystyle v\,du = (-1/2)x^5\sin(x^2)\,dx\ \text{ not } (-1/2)x\sin(x^2)\,dx\ .##

I knew it was something stupid. No idea where I got x from. Thanks Sammy.

## 1. What is integration by parts?

Integration by parts is a method used to evaluate integrals that are in the form of a product of two functions. It is based on the product rule of differentiation and allows us to transform an integral into a simpler form that is easier to evaluate.

## 2. When should I use integration by parts?

Integration by parts is useful when the integral involves a product of functions where one function is easy to integrate but the other is difficult or impossible. In this case, we can choose the easy function to be the first function and the more complicated one to be the second function.

## 3. What is the formula for integration by parts?

The formula for integration by parts is ∫ u dv = u v - ∫ v du, where u and v are the two functions being integrated and du and dv are their respective differentials. This is also known as the integration by parts formula or the integration by parts rule.

## 4. How do I choose which function to be the first function?

When choosing which function to be the first function, it is helpful to use the acronym "LIATE" (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to determine which function is most suitable. In general, we want the first function to become simpler after differentiating.

## 5. Can I use integration by parts more than once?

Yes, it is possible to use integration by parts multiple times if the resulting integral can be simplified. However, it is important to keep track of the terms and to avoid getting caught in an infinite loop. In some cases, it may be more efficient to use other integration techniques rather than repeating integration by parts.