Integration by Parts Twice: How to Solve Tricky Integrals

In summary, the conversation is about solving the integral of e^3x sin x. The person is struggling with the second integration and is looking for guidance. They have made some substitutions and are on the right track, but are having trouble removing the never-ending equations in the integral. The advice given is to do similar substitutions as before and solve the resulting equation algebraically.
  • #1
EthanVandals
55
2

Homework Statement


Integrate e^3x sin x.

Homework Equations


uv - Integral(v du)

The Attempt at a Solution


I am trying to help somebody else with this problem, as I took Calculus a few years ago, but the end is really kicking my butt. I know I'm VERY close, but once I get to the second integration, I just get that u = cos x and du = -sin x. If I assign 1/3 e^3x to be dv, I just seem to get never ending equations in the integral that I can't remove. I know I'm very close, could someone give me a pointer in the right direction?

0IhKC2A.jpg
 
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  • #2
EthanVandals said:

Homework Statement


Integrate e^3x sin x.

Homework Equations


uv - Integral(v du)

The Attempt at a Solution


I am trying to help somebody else with this problem, as I took Calculus a few years ago, but the end is really kicking my butt. I know I'm VERY close, but once I get to the second integration, I just get that u = cos x and du = -sin x. If I assign 1/3 e^3x to be dv, I just seem to get never ending equations in the integral that I can't remove. I know I'm very close, could someone give me a pointer in the right direction?

0IhKC2A.jpg
You're on the right track. Do the same types of substitutions that you did the first time around, and you will have an equation involving the integral you're trying to evaluate. You can solve this equation algebraically.
In short, you will have something like this:
##\int f(x)dx = \text{stuff} + A \int f(x) dx##
##\Rightarrow (1 - A) \int f(x) dx = \text{stuff} + C##
You should be able to solve the last equation for the integral.
 

1. What is Integration by Parts Twice?

Integration by Parts Twice is a method used in calculus to evaluate integrals that cannot be solved using basic integration techniques. It involves using the integration by parts formula multiple times in succession.

2. When should I use Integration by Parts Twice?

Integration by Parts Twice should be used when the integral cannot be solved using simpler methods such as u-substitution or basic integration rules. It is also useful when the integral contains a product of two functions, as this is the type of integral that the integration by parts formula is designed to solve.

3. How does Integration by Parts Twice work?

The integration by parts formula states that the integral of the product of two functions, u and v, can be solved by taking u as the first function and v as the second function, and then using the formula: ∫udv = uv - ∫vdu. Integration by Parts Twice involves using this formula multiple times on the resulting integral until it can be evaluated.

4. What are the steps for using Integration by Parts Twice?

The steps for using Integration by Parts Twice are as follows:

  • 1. Identify the two functions in the integral that can be split into u and v.
  • 2. Take the first function, u, and the second function, v, and use the integration by parts formula: ∫udv = uv - ∫vdu.
  • 3. Repeat the process with the resulting integral, using the same u and v values.
  • 4. Continue this process until the integral can be evaluated.

5. What are some common mistakes made when using Integration by Parts Twice?

One common mistake when using Integration by Parts Twice is not choosing the correct u and v values. It is important to choose u as the function that becomes simpler after differentiation, and v as the function that becomes simpler after integration. Another mistake is not considering the limits of integration, as these may need to be adjusted after each iteration of the formula. Finally, it is important to check your answer by differentiating it to make sure it matches the original integral.

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