Evaluate:
$\displaystyle \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\}$
Answer:
Spoiler:
I do not know how the problem is done. Any ideas?
Let $\displaystyle f(x) = \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\} = \sum_{k=1}^{n} (2k-1) \sin{(2k-1)x}$
Let F(x) one antiderivative of f
Then $\displaystyle F(x) = -\sum_{k=1}^{n} \cos{(2k-1)x}$
$\displaystyle F(x) = -Re\left(\sum_{k=1}^{n} e^{i(2k-1)x}\right)$
After some calculations I can find
$\displaystyle f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}$
But there are many other possible expressions for f(x)
If you use $\displaystyle \sin{(2n+1)x} = \sin(2nx) \cos x + \cos(2nx) \sin x$
you can substitute $\displaystyle \sin(2nx) \cos x = \sin{(2n+1)x} - \cos(2nx) \sin x$
or even $\displaystyle \cos(2nx) \sin x = \sin{(2n+1)x} - \sin(2nx) \cos x$
in the below expression of f
$\displaystyle f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}$
At first I thought that this was the reason why I did not find the same result as yours but then I realized that your solution is not correct