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

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SUMMARY

The integral of (sin 37x / sin x) from 0 to π can be evaluated using the identity for sin(nx) expressed in terms of exponential functions. The formula states that sin(nx) = (1/2i)(X^n - Y^n), where X = e^(ix) and Y = 1/X. This leads to the simplification sin(nx)/sin(x) = X^(n-1) + X^(n-2)Y + ... + Y^(n-1), which can be used to compute the integral directly. The solution involves recognizing the periodic properties of sine functions and leveraging symmetry in the integral bounds.

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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

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

RGV
 

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