∫sin(πt)[cos(2nπt)-sin(2nπt)]dt答案是多少,重分酬谢

Question

∫sin(πt)[cos(2nπt)-sin(2nπt)]dt 答案是多少,重分酬谢

Answer 1

答案：
[-cos（πt+2nπt）]/[2(π+2nπ)]+[-cos（πt-2nπt）]/[2(π-2nπ)]-sin(πt-2nπt)/[2(πt-2nπ)]+sin(πt+2nπt)/[2(πt+2nπ)]
∫sin(πt)[cos(2nπt)-sin(2nπt)]dt
=∫sin(πt)cos(2nπt)dt -∫sin(πt）sin(2nπt)dt
= ∫[sin(πt+2nπt)/2 +sin(πt-2nπt)/2]dt-∫[cos(πt-2nπ)/2-cos(πt+2nπ)/2]
= ∫[sin(πt+2nπt)/2]dt +∫[sin(πt-2nπt)/2]dt-∫[cos(πt-2nπ)/2]dt+∫[cos(πt+2nπ)/2]
=[-cos（πt+2nπt）]/[2(π+2nπ)]+[-cos（πt-2nπt）]/[2(π-2nπ)]-sin(πt-2nπt)/[2(πt-2nπ)]+sin(πt+2nπt)/[2(πt+2nπ)]