leopard
Nov28-08, 02:46 AM
\frac{1}{\pi}\int^{0}_{\pi}x sin(\frac{nx}{2})dx
Attempt at solution:
\frac{1}{\pi} \int^{0}_{\pi}x sin(\frac{nx}{2})dx = \frac{1}{\pi}[\frac {4}{n^2}sin(\frac{nx}{2}) - \frac{2x}{n}cos(\frac{nx}{2})]^{\pi}_0 = \frac{1}{\pi}(\frac{4}{n^2}sin(\frac{n \pi}{2}) - \frac{2 \pi}{n} cos(\frac{n \pi}{2})) = \frac{4}{\pi n^2} sin(\frac{n \pi}{2}) -\frac{2}{n}cos(\frac{n \pi}{2})
The correct answer according to the solutions manual is
\frac{4}{\pi n^2} sin(\frac{n \pi}{2}) - \frac{1}{\pi}cos(\frac{n \pi}{2})
Where's my mistake?
Attempt at solution:
\frac{1}{\pi} \int^{0}_{\pi}x sin(\frac{nx}{2})dx = \frac{1}{\pi}[\frac {4}{n^2}sin(\frac{nx}{2}) - \frac{2x}{n}cos(\frac{nx}{2})]^{\pi}_0 = \frac{1}{\pi}(\frac{4}{n^2}sin(\frac{n \pi}{2}) - \frac{2 \pi}{n} cos(\frac{n \pi}{2})) = \frac{4}{\pi n^2} sin(\frac{n \pi}{2}) -\frac{2}{n}cos(\frac{n \pi}{2})
The correct answer according to the solutions manual is
\frac{4}{\pi n^2} sin(\frac{n \pi}{2}) - \frac{1}{\pi}cos(\frac{n \pi}{2})
Where's my mistake?