用換元法
令t=x-pi/2
則原試變?yōu)椋骸?pi/2到pi/2
(t+pi/2)sin(t+pi/2)/1+cos(t+pi/2)^2 dt
=∫-pi/2到pi/2 (t+pi/2)cost/1+sint^2 dt
=∫-pi/2到pi/2 tcost/1+(sint)^2 dt
+ ∫-pi/2到pi/2 pi/2 *cost/1+sint^2 dt
顯然 tcost/1+(sint)^2為奇函數(shù) 故他的積分為零
而∫-pi/2到pi/2
pi/2*cost/1+sint^2 dt
=pi/2 ∫-pi/2到pi/2 1/1+sint^2 d sint
=pi/2 *( arctan(sint))-pi/2到pi/2
=pi/2 * (pi/4+pi/4)
=pi^2/4
即原式=pi^2/4