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

• dangsy
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.
dangsy

## Homework Statement

$$\int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx$$

## The Attempt at a Solution

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

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

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

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

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

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

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

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

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

$$Sin\left(2n\pi\right) = 0$$
$$Sin\left(2na\pi\right) = 0$$

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

I'm kinda stuck here...

I know the answer is $$\frac{a}{2}$$

but I'm not sure how to get to it...maybe an integration mistake?

first of all you are claiming that

$$\frac{d}{dx}\left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right]=Sin^2\left(\frac{nx\pi}{a}\right)$$ which is not true

You need to use the identity $$\cos(2x) = 1 - 2\sin^2(x)$$ to turn $$\sin^2(x)$$ 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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