Can You Prove the Reduction Formula for Integral of Sin(nx) Over Sin(x)?

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If I n =integral sin (nx) dx/sinx,prove that In =2sin((n-1)x)/n-1 +In-2 for all integers n>(egual) .


I don't know how to start...
 
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Since you are given the formula, the obvious approach is to use induction.
 
e179285 said:
If I n =integral sin (nx) dx/sinx,prove that In =2sin((n-1)x)/n-1 +In-2 for all integers n>(equal) .

I don't know how to start...
sin(nx) = sin((n-1)x+x)
=sin((n-1)x)∙cos(x) + cos((n-1)x)∙sin(x)​

is a good place to start.
 
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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