b
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 May 23, 2009 #2 −π∫02t2sin(πt)dt\displaystyle {-\pi}\int_{0}^{2}t^{2}sin({\pi}t)dt−π∫02t2sin(πt)dt Did you try using parts?. Let u=t2, dv=sin(πt)dt, du=2tdt, v=−cos(πt)π\displaystyle u=t^{2}, \;\ dv=sin({\pi}t)dt, \;\ du=2tdt, \;\ v=\frac{-cos({\pi}t)}{\pi}u=t2, dv=sin(πt)dt, du=2tdt, v=π−cos(πt) The second integral evaluates to 0, so don't worry about it. Just do this one.
−π∫02t2sin(πt)dt\displaystyle {-\pi}\int_{0}^{2}t^{2}sin({\pi}t)dt−π∫02t2sin(πt)dt Did you try using parts?. Let u=t2, dv=sin(πt)dt, du=2tdt, v=−cos(πt)π\displaystyle u=t^{2}, \;\ dv=sin({\pi}t)dt, \;\ du=2tdt, \;\ v=\frac{-cos({\pi}t)}{\pi}u=t2, dv=sin(πt)dt, du=2tdt, v=π−cos(πt) The second integral evaluates to 0, so don't worry about it. Just do this one.
