# Calculus Simple Integration by Parts question

• koolkai5953
In summary, the problem is finding the antiderivative from 0 to 1 of (e^x)*sin(Nx)dx. The person has attempted to integrate by parts multiple times, but is unsure if they are doing it correctly. They are also unsure if they are supposed to substitute sin(Nx) for something else. The answer must include the constant N, which can be replaced with any number. The solution involves looking at the problem algebraically and using the fact that integrating by parts twice should result in the original integral. The person is grateful for the assistance and notes that calculus is like a puzzle that requires an open mind to solve.
koolkai5953
1. The problem is this:

The antiderivative from 0 to 1 of (e^x)*sin(Nx)dx

I tried integrating by parts several times and I'm just not sure if what I'm doing was correct. I keep hitting a dead end. I'm not sure if I'm supposed to IBP twice or substitute sin(Nx) for something else. Help please :) The rest of the questions rely on this answer so please and thank you for any assistance. O yea, and the answer has to have N in it. N is just a constant that can be replaced with any number. thanks again

This is a case where no matter what you choose to be your u and dv, the terms will never reach zero and it will seem that you have to do it forever. However, you can look at it from an algebraic standpoint. When you perform integration by parts twice, you should end up with the integral of what you started with. You can add only this integral part to the left hand side of your equation to get 2[(e^x)*sin(Nx)dx]. You then can evaluate whatever is left on the right side of the equation from 0 to 1 and divide by two. This will give you the value of your initial integral.

$$I am \left( \int_0^1 e^x \cdot e^{iNx}\ \mbox{d}x \right)$$

thank you very much for your quick response. i was on the right track but i just didnt see the algebraic part. calculus is like a giant puzzle! u have to look at these problems with an open mind. take care and have a nice day
-kyle

## What is integration by parts?

Integration by parts is a method in calculus used to find the integral of a product of two functions. It involves breaking down a complicated integral into simpler parts that can be more easily integrated.

## When is integration by parts used?

Integration by parts is used when the integral of a function cannot be found using other methods, such as substitution or trigonometric identities. It is also commonly used when the integral involves products of functions.

## What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are functions and du and dv represent their differentials.

## How do you choose u and dv for integration by parts?

When choosing u and dv, it is important to follow the acronym "LIATE", which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. Generally, u should be the function that appears earlier in the acronym and dv should be the function that appears later.

## What is the process for solving a simple integration by parts question?

The process for solving a simple integration by parts question involves choosing u and dv, applying the formula ∫u dv = uv - ∫v du, simplifying the integral, and then solving for the original integral. It may also involve multiple rounds of integration by parts if the integral is still not easily solvable.

• Calculus and Beyond Homework Help
Replies
2
Views
408
• Calculus and Beyond Homework Help
Replies
3
Views
983
• Calculus and Beyond Homework Help
Replies
7
Views
955
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
534
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K