How to solve an integral of absolute value of sin finction ??

[pisrationalhahaha]

\(\displaystyle \int_{0}^{2\pi } \left | sin(\frac{n\theta -\theta }{2}) \right |d\theta\)



How can we solve this integral without using correction factors thing (I don't know what is this)(searched for it on web)

 

[Dr.Peterson]

\(\displaystyle \int_{0}^{2\pi } \left | sin(\frac{n\theta -\theta }{2}) \right |d\theta\)

How can we solve this integral without using correction factors thing (I don't know what is this)(searched for it on web)

Can you tell us about the context of the problem? Are you told that n is an integer, for example? And what topics have you learned recently that it might be intended to test? Is the "correction factors thing" something you found somewhere online, or part of the problem?

If n is an integer, then you are just integrating over some number of whole or half periods, and you can find the integral over one period (or 1/2 period) as a first step.

 

[pisrationalhahaha]

Can you tell us about the context of the problem? Are you told that n is an integer, for example? And what topics have you learned recently that it might be intended to test? Is the "correction factors thing" something you found somewhere online, or part of the problem?

If n is an integer, then you are just integrating over some number of whole or half periods, and you can find the integral over one period (or 1/2 period) as a first step.

n is a natural number greater than or equal to 2 (sorry for missing this)
There are no topics that I have learned recently that could be related
" correction factors " I've found it online
How to solve it ?

 

[Dr.Peterson]

n is a natural number greater than or equal to 2 (sorry for missing this)
There are no topics that I have learned recently that could be related
" correction factors " I've found it online
How to solve it ?

I have no idea what you found online, or what you searched for. But you can ignore it.

Have you tried following up my suggestion, since I was right about the integer?

What is the period of the function, for a given n? Sketch the graph; what is the area of each "hump"? How many are there?

 

[pisrationalhahaha]

I have no idea what you found online, or what you searched for. But you can ignore it.

Have you tried following up my suggestion, since I was right about the integer?

What is the period of the function, for a given n? Sketch the graph; what is the area of each "hump"? How many are there?

Okay, I tried to do it this way :
we know that
\(\displaystyle 0\leq \theta \leq 2\pi\)

\(\displaystyle \Rightarrow 0\leq (\frac{n-1}{2})\theta \leq \pi (n-1)\)

Let \(\displaystyle U=(\frac{n-1}{2})\theta\) \(\displaystyle \Rightarrow\) \(\displaystyle \frac{dU}{d\theta }=\frac{n-1}{2}\)

So \(\displaystyle \int_{0}^{2\pi}\left | sin(\frac{n-1}{2}\theta ) \right |d\theta \Leftrightarrow \int_{0}^{(n-1)\pi }2\frac{\left | sin(U) \right |}{n-1}dU\)

\(\displaystyle \int_{0}^{(n-1)\pi }2\frac{\left | sin(U) \right |}{n-1}dU=\frac{2}{n-1}(\int_{0}^{\pi}sin(U)dU+\int_{\pi}^{2\pi}-sin(U)dU+\int_{2\pi}^{3\pi}sin(U)dU+...............+\int_{(n-2)\pi}^{(n-1)\pi}\pm sin(U)dU)\) (all of these definite integrals are equal to 2)

\(\displaystyle \Leftrightarrow (\frac{2}{n-1})\sum_{1}^{n-1}(2)=(\frac{2}{n-1})(\frac{2+2}{2})(n-1)=4\)

 

[Dr.Peterson]

Okay, I tried to do it this way :
we know that
\(\displaystyle 0\leq \theta \leq 2\pi\)

\(\displaystyle \Rightarrow 0\leq (\frac{n-1}{2})\theta \leq \pi (n-1)\)

Let \(\displaystyle U=(\frac{n-1}{2})\theta\) \(\displaystyle \Rightarrow\) \(\displaystyle \frac{dU}{d\theta }=\frac{n-1}{2}\)

So \(\displaystyle \int_{0}^{2\pi}\left | sin(\frac{n-1}{2}\theta ) \right |d\theta \Leftrightarrow \int_{0}^{(n-1)\pi }2\frac{\left | sin(U) \right |}{n-1}dU\)

\(\displaystyle \int_{0}^{(n-1)\pi }2\frac{\left | sin(U) \right |}{n-1}dU=\frac{2}{n-1}(\int_{0}^{\pi}sin(U)dU+\int_{\pi}^{2\pi}-sin(U)dU+\int_{2\pi}^{3\pi}sin(U)dU+...............+\int_{(n-2)\pi}^{(n-1)\pi}\pm sin(U)dU)\) (all of these definite integrals are equal to 2)

\(\displaystyle \Leftrightarrow (\frac{2}{n-1})\sum_{1}^{n-1}(2)=(\frac{2}{n-1})(\frac{2+2}{2})(n-1)=4\)

That's not quite the direction I suggested, but is a very nice formal alternative. It looks good.

 

[pisrationalhahaha]

That's not quite the direction I suggested, but is a very nice formal alternative. It looks good.

Can you solve it using the method you suggested please ?

 

[Dr.Peterson]

Can you solve it using the method you suggested please ?

Yours is actually better, and easier to be sure it's right. Essentially, what I did was to observe that you are integrating over n-1 periods of the absolute value of the sine (the period being the interval in which the argument goes from 0 to pi). Then I would carry out the integral much as you did, but only from 0 to 2 pi/(n-1), and multiply by n-1. Your work did the same thing; mine just avoids the need for your summation.