sin2x和cos2x的原函数求解过程 ，谢谢
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绛旓細璇
绛旓細鍥犱负((sin(2x))/2)'=cos2x锛屾墍浠(sin(2x))/2鏄痗os2x鐨勫師鍑芥暟銆備緥寮忓涓嬶細鈭玞os2xdx =1/2鈭玞os2xd2x =(sin2x)/2+C 鍘熷嚱鏁帮紙primitive function锛夋槸鎸囧浜庝竴涓畾涔夊湪鏌愬尯闂寸殑宸茬煡鍑芥暟f(x)锛屽鏋滃瓨鍦ㄥ彲瀵煎嚱鏁癋(x)锛屼娇寰楀湪璇ュ尯闂村唴鐨勪换涓鐐归兘瀛樺湪dF(x)=f(x)dx锛屽垯鍦ㄨ鍖洪棿鍐呭氨绉板嚱鏁...
绛旓細(sin2x)' = 2cos2x sin2x鐨勫師鍑芥暟鏄(-1/2)cos2x + C锛屽鏁版槸2cos2x銆侰涓哄父鏁般傦紙4锛夆埆cos2xdx = (1/2)鈭玞os2xd(2x) = (1/2)sin2x + C (cos2x)' = -sin2x * 2 = -2sin2x cos2x鐨勫師鍑芥暟鏄(1/2)sin2x + C锛屽鏁版槸-2sin2x銆侰涓哄父鏁般
绛旓細瑙ｏ細(sin2x)'=cos(2x)路(2x)'=2cos(2x)鍑芥暟sin(2x)鏄嚱鏁癴(x)=2cos(2x)鐨勫師鍑芥暟銆傝В棰樻濊矾锛1銆佽嫢d[F(x)]=f(x)dx锛屽垯F(x)鏄痜(x)鐨勫師鍑芥暟銆傚彧闇瀵箂in(2x)姹傚锛屽嵆鍙緱鍒版墍姹傜殑鍑芥暟銆2銆佹眰瀵艰繃绋嬩腑鐢ㄥ埌鐨勫叕寮忥細(sinx)'=cosx (Cx)'=C锛(鍏朵腑锛孋涓哄父鏁帮紝鏈涓紝C=2)3銆...
绛旓細sin2x鐨勫師鍑芥暟F锛坸)=鈭玸in2xdx=鈭(1/2)sin2xd2x=(1/2)鈭玠(-cos2x)=-1/2cos2x
绛旓細鍥犱负((sin(2x))/2)'=cos2x锛鎵浠(sin(2x))/2鏄痗os2x鐨勫師鍑芥暟銆備緥寮忓涓嬶細鈭玞os2xdx =1/2鈭玞os2xd2x =(sin2x)/2+C 渚嬪 x3鏄3x2鐨勪竴涓師鍑芥暟锛屾槗鐭ワ紝x3+1鍜寈3+2涔熼兘鏄3x2鐨勫師鍑芥暟銆傚洜姝わ紝涓涓嚱鏁板鏋滄湁涓涓師鍑芥暟锛屽氨鏈夎璁稿澶氬師鍑芥暟锛屽師鍑芥暟姒傚康鏄负瑙ｅ喅姹傚鍜屽井鍒嗙殑閫嗚繍绠楄...
绛旓細鍥犱负((sin(2x))/2)'=cos2x锛鎵浠(sin(2x))/2鏄痗os2x鐨勫師鍑芥暟銆備緥寮忓涓嬶細鈭玞os2xdx =1/2鈭玞os2xd2x =(sin2x)/2+C 涓嶅畾绉垎鐨勫叕寮 1銆佲埆 a dx = ax + C锛宎鍜孋閮芥槸甯告暟 2銆佲埆 x^a dx = [x^(a + 1)]/(a + 1) + C锛屽叾涓璦涓哄父鏁颁笖 a 鈮 -1 3銆佲埆 1/x dx =...
绛旓細鍘熷紡=鈭玞ot²xdx =鈭(csc²-1)dx =-cotx-x+C
绛旓細鍘熷嚱鏁=鈭cos2xdx =1/2鈭玞os2xd2x =1/2sin2x+C
绛旓細cos2x鐨勫師鍑芥暟鏄1/2*sin2x+C锛孋涓哄父鏁般傝В锛氬師鍑芥暟鍙互閫氳繃涓嶅畾绉垎鏉ユ眰鍙栥備护f(x)=cos2x锛宖(x)鐨勫師鍑芥暟涓篎(x)锛屽垯F(x)=鈭玣(x)dx銆傞偅涔團(x)=鈭玣(x)dx=鈭玞os2xdx=1/2*鈭玞os2xd(2x)=1/2*sin2x+C锛孋涓哄父鏁般傚嵆cos2x鐨勫師鍑芥暟鏄1/2*sin2x+C锛孋涓哄父鏁般備笉瀹氱Н鍒嗗噾寰垎娉...
