I have to calculate the following integral $$ b_n = \dfrac{1}{\pi} \int_{-\pi}^{\pi} \dfrac{1}{2}x \sin nx dx$$
The correct answer is apparently $$\dfrac{(-1)^{n-1}}{n}$$
But I have no idea how I should get this answer. I was able to find that $$\int \dfrac{1}{2}x\sin nx dx = \dfrac{\sin nx - nx \cos nx}{2n^2}$$
How should I proceed?
\begin{eqnarray*} f(\alpha ) &=&\frac{1}{2\pi }\int_{-\pi }^{+\pi }dxx\sin (\alpha x)=-\frac{1% }{2\pi }\partial _{\alpha }\int_{-\pi }^{+\pi }dx\cos (\alpha x)=-\frac{1}{% 2\pi }\partial _{\alpha }\left( \frac{\sin (\alpha x)}{\alpha }\right) _{x=-\pi }^{+\pi } \\ &=&-\frac{1}{2\pi }\partial _{\alpha }\left( \frac{\sin (\alpha \pi )}{% \alpha }-\frac{\sin (-\alpha \pi )}{\alpha }\right)\newline&=&-\frac{1}{\pi }\partial _{\alpha }\left( \frac{\sin (\alpha \pi )}{\alpha }\right) =\frac{1}{\pi }% \left( \frac{\sin (\alpha \pi )}{\alpha ^{2}}-\pi \frac{\cos (\alpha \pi )}{% \alpha }\right) \end{eqnarray*} $$ b_{n}=f(n)=-\frac{\cos (n\pi )}{n}=(-1)^{n-1}\frac{1}{n} $$