求不定积分∫(cos^4x)dx,
先降次把cos^4x降为cos^2x*cos^2x再把cos^2x降为1/2(cos2x+1)由于有两项这个式子相乘次数又升高了再次用倍角公式降次降到一次为止别忘了c绛旓細浣犲ソ锛佸埄鐢ㄥ嶈鍏紡闄嶆鍐绉垎 璇︾粏瑙ｇ瓟濡傚浘 婊℃剰璇峰ソ璇 璋㈣阿o(鈭鈭)o
绛旓細=(1/4)鈭玔1-2cos2x+(cos2x)^2]dx =(1/4)鈭玔1-2cos2x+(1/2)(1+cos4x)]dx =(3/8)鈭玠x-(1/2)鈭玞os2xdx+(1/8)鈭玞os4xdx =(3/8)鈭玠x-(1/4)鈭玞os2xd2x+(1/32)鈭玞os4xd4x =(3/8)x-(1/4)sin2x+(1/32)sin4x+C 涓涓嚱鏁帮紝鍙互瀛樺湪涓嶅畾绉垎锛岃屼笉瀛樺湪瀹氱Н鍒嗭紝...
绛旓細璁緐=4x锛寈=u/4锛宒x=1/4du銆傚師寮=1/4鈭玞osudu =1/4sinu+C =1/4sin4x+C 鎵浠cos4xdx鐨涓嶅畾绉垎鏄1/4sin4x+C銆備笉瀹氱Н鍒嗙殑鍏紡 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 = l...
绛旓細(cosx)^4 =cos⁴x =(cos²x)²=[(1+cos2x)/2]²=(1/4)(1+2cos2x+cos²2x)=(1/4)+(1/2)cos2x+(1/8)(1+cos4x)=(3/8)+(1/2)cos2x+(1/8)cos4x鈭cos⁴xdx =鈭玔(3/8)+(1/2)cos2x+(1/8)cos4x]dx =(3/8)x+(1/4)sin2x+...
绛旓細鈭 cos^4x dx = 鈭 (cos²x)² dx = 鈭 [1/2*(1+cos2x)]² dx = (1/4)鈭 (1+2cos2x+cos²2x) dx = (1/4)鈭 dx + (1/4)鈭 cos2x d(2x) + (1/4)鈭 1/2*(1+cos4x) dx锛岃嫢瑕佽鎹㈠厓娉曪紝杩欓噷鍙互鐢 = (1/4)x + (1/4)sin2x + (1...
绛旓細鍒╃敤cos鈭2x锛½锛1-cos2x锛夋妸cos鈭4x鍖栫畝鎴3/8-½cos2x-1/8cos4x,鐒跺悗姹傜Н鍒銆傚緱3x/8-¼sin2x-sin4x/32+c 璇濊鍒嗙被濂藉儚鏀鹃敊浜嗐傘傘
绛旓細=(1/4)(1+2cos2x+cos²2x)=(1/4)+(1/2)cos2x+(1/8)(1+cos4x)=(3/8)+(1/2)cos2x+(1/8)cos4x鈭玞os⁴xdx =鈭玔(3/8)+(1/2)cos2x+(1/8)cos4x]dx =(3/8)x+(1/4)sin2x+(1/32)sin4x+C 涓嶅彲绉嚱鏁 铏界劧寰堝鍑芥暟閮藉彲閫氳繃濡備笂鐨勫悇绉嶆墜娈佃绠楀叾涓嶅畾绉...
绛旓細cos2x)^2dx =x/4+C+1/4鈭玞os2xd(2x)+1/4鈭玔(1+cos4x)/2]dx =x/4+(sin2x)/4+C+1/4鈭1/2dx+1/4鈭(cos4x)/2dx =3x/8+(sin2x)/4+C+1/32鈭4cos4xdx =3x/8+(sin2x)/4+C+1/32鈭玞os4xd(4x)=3x/8+(sin2x)/4+(sin4x)/32+C ...
绛旓細I=(1/4)鈭(cos2x+1)^2dx锛堝嶈鍏紡锛=(1/4)鈭(cos2x)^2dx+(1/2)鈭玞os2xdx+(1/4)鈭玠x锛堝睍寮锛=(1/8)鈭(cos4x)dx+(1/2)鈭玞os2xdx+3/8鈭玠x锛堝嶈鍏紡+鍚堝苟鍚岀被椤癸級=(1/32)sin4x+(1/4)sin2x+(3x/8)+C锛堝噾寰垎娉曪級
绛旓細涓ゆ鐢ㄥ叕寮忋恈os²x=(1+cos2x)/2銆戝緱鍒cos^4=(cos²)²=(1+cos2x)²/4=鈥︺
