Integrating e^(2 theta)sin(9 theta) d theta

In summary, the student attempted to integrate by parts and then Integrated by Parts again, but got the same answer. After simplifying the answer, they found that they had made an arithmetic or algebra error.
  • #1
tangibleLime
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0

Homework Statement



Integrate.

[tex]\int e^{2\theta}sin(9\theta) d\theta[/tex]​

Homework Equations



Integration by Parts

[tex]\int udv = uv - \int vdu[/tex]​

The Attempt at a Solution



I started out with integration by parts (IPB).

[tex]u = e^{2\theta}[/tex]

[tex]du = 2e^{2\theta} d\theta[/tex]

[tex]v = \frac{-1}{9}cos(9\theta)[/tex]

[tex]dv = sin(9\theta) d\theta[/tex]

[tex]uv - \int vdu = -\frac{1}{9}cos(9\theta)*e^{2\theta}+\frac{2}{9}\int cos(9\theta) d\theta[/tex]​
Noticing that the new integral on the right is just about the same as how I started (sin just turned into cos), I did IPB once again.

[tex]u = e^{2\theta}[/tex]

[tex]du = 2e^{2\theta} d\theta[/tex]

[tex]v = \frac{1}{9}sin(9\theta}[/tex]

[tex]dv = cos(9\theta) d\theta[/tex]

[tex]uv - \int vdu = \frac{1}{9}sin(9\theta)*e^{2\theta}-\frac{2}{9}\int sin(9\theta)*e^{2\theta} d\theta[/tex]​
Combining all of this into the equation so far...

[tex]-\frac{1}{9}cos(9\theta)*e^{2\theta}+\frac{2}{81}sin(9\theta)*e^{2\theta}-\frac{2}{9}\int sin(9\theta)*e^{2\theta} d\theta[/tex]​
Noticing that this new integral on the right is the exactly equal to what I started with, I added it to both sides in order to remove it from the right side. I mentally noted that the left version could have a coefficient of [tex]\frac{9}{9}[/tex], and added the right side's coefficient of [tex]\frac{2}{9}[/tex] to give me the final coefficient for the left version of [tex]\frac{11}{9}[/tex].

[tex]\frac{11}{9}\int e^{2\theta}sin(9\theta) d\theta = -\frac{1}{9}cos(9\theta)e^{2\theta}+\frac{2}{81}sin(9\theta)e^{2\theta}[/tex]​
To get that [tex]\frac{11}{9}\[/tex] out of the left side of the equation to get my original integral back, I multiplied both sides by it's reciprocal, [tex]\frac{9}{11}[/tex]. Then I tacked on the constant of integration to arrive at my final answer.

[tex]\int e^{2\theta}sin(9\theta) d\theta = \frac{9}{11}(-\frac{1}{9}cos(9\theta)e^{2\theta}+\frac{2}{81}sin(9\theta)e^{2\theta}) + C[/tex]

This was marked wrong, even after I simplified the answer (which I did not do here, but I checked the simplification with WolframAlpha). Any suggestions?​
 
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  • #2
If you differentiate your result, you get 85/99 e2tsin 9t, which is almost what you started with, so it looks like you just made an arithmetic or algebra error somewhere. Your method is correct.
 
  • #3
Thanks, I figured as much, that's usually my downfall :D As long as I know my method is correct, I'm okay with it.
 

Related to Integrating e^(2 theta)sin(9 theta) d theta

1. What is the purpose of integrating e^(2 theta)sin(9 theta) d theta?

The purpose of integrating e^(2 theta)sin(9 theta) d theta is to find the area under the curve of the function e^(2 theta)sin(9 theta) with respect to the variable theta. This is useful in many areas of science, including physics and engineering, as it allows for the calculation of important quantities such as work and displacement.

2. What are the steps to integrate e^(2 theta)sin(9 theta) d theta?

The steps to integrate e^(2 theta)sin(9 theta) d theta are as follows:

  1. Use the product rule to rewrite the function as e^(2 theta)(1/2)(-cos(9 theta)), which can then be simplified to e^(2 theta)cos(9 theta).
  2. Apply the power rule for integration by raising the power of e to (2 theta + 1) and dividing by the new power.
  3. Apply the chain rule for integration by substituting u = 9 theta and du = 9 d theta in the integral.
  4. Integrate the new function by using the power rule for integration again.
  5. Substitute back in u and simplify the final result.

3. Can e^(2 theta)sin(9 theta) d theta be integrated using a u-substitution?

Yes, e^(2 theta)sin(9 theta) d theta can be integrated using a u-substitution. The appropriate substitution is u = 9 theta, which will result in the integral becoming e^(2 theta)(1/2)(-cos(u))du, which can then be solved using the power rule for integration.

4. What is the result of integrating e^(2 theta)sin(9 theta) d theta?

The result of integrating e^(2 theta)sin(9 theta) d theta is (1/2)e^(2 theta)sin(9 theta) - (1/18)e^(2 theta)cos(9 theta) + C, where C is the constant of integration.

5. How can the integral of e^(2 theta)sin(9 theta) d theta be used in real-world applications?

The integral of e^(2 theta)sin(9 theta) d theta can be used in real-world applications to calculate the work done by a force acting on an object, or the displacement of an object under a given force. It can also be used in physics and engineering to solve problems involving simple harmonic motion, as well as in other areas of science such as electricity and magnetism.

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