Solving Integration Problem: \int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx

In summary: Secondly, you are saying that\left| x \left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right] \right|^{a}_{0}- \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}This is incorrect. You need to use the product rule to integrate this.
  • #1
dangsy
15
0

Homework Statement



[tex]\int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx[/tex]

The Attempt at a Solution


[tex]\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\, dx\[/tex]

[tex]\left| x \left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right] \right|^{a}_{0}- \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}[/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}[/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a} \int_0^a nx\pi-\frac{Sin\left(2nx\pi\right)}{2}[/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ \int_0^a nx\pi-\int_0^a\frac{Sin\left(2nx\pi\right)}{2}\right][/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi\int_0^a x- \frac{1}{2}\int_0^a Sin\left(2nx\pi\right)\right][/tex]


[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi \left | x\right |^{a}_{0} - \frac{1}{2} \left| Sin\left(2nx\pi\right)\right|^{a}_{0}\right][/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi a - \frac{1}{2} Sin\left(2na\pi\right)\right][/tex]

[tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \left[ \frac{n\pi}{2} - \frac{1}{4a} Sin\left(2na\pi\right)\right][/tex]

[tex]Sin\left(2n\pi\right) = 0[/tex]
[tex]Sin\left(2na\pi\right) = 0 [/tex]

[tex]a \left[ \frac{n\pi}{2}\right] - \left[ \frac{n\pi}{2} \right][/tex]

I'm kinda stuck here...

I know the answer is [tex]\frac{a}{2}[/tex]

but I'm not sure how to get to it...maybe an integration mistake?
 
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  • #2
first of all you are claiming that

[tex]\frac{d}{dx}\left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right]=Sin^2\left(\frac{nx\pi}{a}\right)[/tex] which is not true
 
  • #3
You need to use the identity [tex]\cos(2x) = 1 - 2\sin^2(x)[/tex] to turn [tex]\sin^2(x)[/tex] into a sum of things you know how to integrate (either directly or using integration by parts).
 

1. What is integration and why is it important in science?

Integration is a mathematical process that involves finding the area under a curve. In science, integration is important because it allows us to calculate important quantities such as displacement, velocity, and acceleration from data collected in experiments.

2. How do I solve the integration problem \int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx?

To solve this integration problem, you can use the trigonometric identity Sin^2(x) = (1/2)(1-cos(2x)), which will simplify the integral to \frac{a^2}{4n\pi^2} - \frac{a}{2n\pi}cos(n\pi). Then, you can use integration by parts to solve for the cosine term and evaluate the integral for the first term.

3. What is the significance of the limits of integration, 0 and a, in this problem?

The limits of integration, 0 and a, represent the starting and ending points of the integration. In this problem, the limits are used to indicate that we are finding the area under the curve from x=0 to x=a.

4. How does the value of n affect the solution to this integration problem?

The value of n affects the solution to this integration problem by changing the frequency of the sine function. This results in different values for the area under the curve and the final solution.

5. Can this integration problem be solved using other methods?

Yes, this integration problem can also be solved using substitution or the trigonometric identity Sin^2(x) = (1/2)(1-cos(2x)). However, the method used may depend on the specific problem and the individual's preference.

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