Can You Help Me with This Integration Problem?

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The discussion revolves around integrating the function sin(2nx)/sin(x) for n=1,2,3, and finding a way to express sin(2nx) as a product involving sin(x). Participants suggest using exponential forms of sine to facilitate the integration process, which can lead to a sum of trigonometric functions. The conversation highlights the use of trigonometric identities to express sin(nx) and cos(nx) in terms of sin(x) and cos(x). There is a focus on understanding how to manipulate these identities to achieve the desired representation for integration. The thread emphasizes the need for clarity in transforming these functions to solve the integration problem effectively.
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Can you help me with
\int\frac{\sin(2nx)}{\sin(x)}dx
Here n=1,2,3...
I think that i should get any way to represent \sin(2nx) as product of sinx and something. But i don't know how.
Thank you
 
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Except for the integration constant,here's what Mathematica gives as an answer.

Daniel.
 

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Great. I have Mathematica too.
I'm given a hint. sin(2nx)=sin(x)*(Sum of trigonometric functions). I don't even understand how my head had to work to get such an idea.
How i had to think about this problem??
 
What are those equal to...?

\sin nx =...?

\cos nx =...?

in terms of the powers of "sin" and "cos" of "x"...?

Daniel.
 
To write it as sin(x)*(Sum of trigonometric functions) you can replace you sines with exponentials, that is \sin(y)=(e^{iy}-e^{-iy})/(2i). Things will factor, and you should be able to pull out a sum of cosines.
 
:blushing: I know only

\sin nx =\sin x \cos[(n-1)x] + \cos x \sin[(n-1)x]
\cos nx =\cos x \cos[(n-1)x] - \sin x \sin[(n-1)x]

These transformations can be maid also with \sin[(n-1)x], and so on.
But how can i write that as a sum?
 
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