How Do You Solve the Integral of (sin 37x / sin x) from 0 to π?

jd12345
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Homework Statement


Find the value of ∫ ( sin 37x / sin x )dx - from 0 to π


Homework Equations





The Attempt at a Solution


Well i tried doing it by parts but it makes its more complicated. I tried to expand sin 37x as sin(36+1)x but could not solve it. I am lost
 
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jd12345 said:

Homework Statement


Find the value of ∫ ( sin 37x / sin x )dx - from 0 to π


Homework Equations





The Attempt at a Solution


Well i tried doing it by parts but it makes its more complicated. I tried to expand sin 37x as sin(36+1)x but could not solve it. I am lost

\sin(nx) = \frac{1}{2i} ( X^n - Y^n), \text{ where } X = e^{ix}, \: Y = 1/X.
Thus, \sin(nx) = \sin(x) [X^{n-1} + X^{n-2}Y + \cdots + Y^{n-1}], so
\sin(nx)/\sin(x) = X^{n-1} + X^{n-2}Y + \cdots + Y^{n-1}.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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