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I have to integrate this expression so I started to solve the delta part from the fact that when n=0 it results equals to 1.

And the graph is continuous in segments I thought as the sumation of integers

$$ \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx $$

From the fact that actually

$$ δ(sin(x))= \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$

An in the integral

$$ \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$

For n=0

$$ \int_{-(1/2)π}^{(1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ) = 1 $$

But taking a look on the graph of 1/|Cos(x)| and the fact that it is continuous in segments

for n=0 ,1 segment

n=1, 3 segments

n=2, 5 segments

and each one of these is just like the first whose integral is equal to one.

So the value it just a sumation that depends for the n value.

And the result is

$$ \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx= \sum_{N=0}^n (2N+1)$$

Has anyone solved it before? Is this a correct way of doing it?

Thanks for your advise.

Jorge M

And the graph is continuous in segments I thought as the sumation of integers

$$ \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx $$

From the fact that actually

$$ δ(sin(x))= \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$

An in the integral

$$ \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$

For n=0

$$ \int_{-(1/2)π}^{(1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ) = 1 $$

But taking a look on the graph of 1/|Cos(x)| and the fact that it is continuous in segments

for n=0 ,1 segment

n=1, 3 segments

n=2, 5 segments

and each one of these is just like the first whose integral is equal to one.

So the value it just a sumation that depends for the n value.

And the result is

$$ \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx= \sum_{N=0}^n (2N+1)$$

Has anyone solved it before? Is this a correct way of doing it?

Thanks for your advise.

Jorge M

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