求sin∧2xcosx不定积分帮忙作业
\u222bcosx/sin^2x\u4e0d\u5b9a\u79ef\u5206\u222bcos(x)/(sin(x)^2)dx
=\u222b1/(sin(x)^2)dsin(x)
=-1/sin(x)+C
\u222bdx/(sin2xcosx)=\u222bdx/(2sinxcos²x)=\u222b(1/(2sinx(1-sin²x))dx=1/2\u222b[1/sinx + sinx/(1-sin²x)]dx=1/2\u222b(cscx+sinx/cos²x)dx=1/2\u222bcscxdx-1/2\u222b1/cos²x d(cosx)=1/2*ln|cscx-cotx|+1/2*secx+C
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绛旓細=1/4鈭玿sin2xd2x =-1/4鈭玿dcos2x =xcos2x/4+1/4鈭玞os2xdx =-xcos2x/4+sin2x/8+C 涓嶅畾绉垎鐨勬剰涔夛細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鍒嗭紝涔熷彲浠ュ瓨鍦ㄥ畾绉垎锛岃屾病鏈変笉瀹氱Н鍒嗐傝繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜屼笉瀹氱Н鍒嗐傝嫢鍦ㄦ湁闄愬尯闂碵a,b]涓婂彧鏈夋湁闄愪釜闂存柇鐐逛笖鍑芥暟鏈夌晫锛鍒瀹氱Н鍒...
绛旓細=-(1/鈭5+1锛塴n|鈭5/2-(cosx-1/2)|+(1/鈭5-1)ln|鈭5/2+(cosx-1/2)|+c =(1/鈭5-1)ln|鈭5/2+(cosx-1/2)|-(1/鈭5+1锛塴n|鈭5/2-(cosx-1/2)|+c 杩炵画鍑芥暟锛屼竴瀹氬瓨鍦ㄥ畾绉垎鍜涓嶅畾绉垎锛涜嫢鍦ㄦ湁闄愬尯闂碵a,b]涓婂彧鏈夋湁闄愪釜闂存柇鐐逛笖鍑芥暟鏈夌晫锛鍒瀹氱Н鍒嗗瓨鍦紱鑻ユ湁璺宠穬銆...
绛旓細=鈭1/[2sinx(cosx)^2]dx = 鈭玔(sinx)^2+(cosx)^2]/[2sinx(cosx)^2]dx =1/2鈭玸ecxtanxdx+1/2鈭玞scxdx =1/2tanx+1/2ln|cscx-cotx|+C
绛旓細鈭cosx/(4-sin^2x)dx =鈭1/(4-sin^2x)dsinx =1/4鈭玔1/(2-sinx)+1/(2+sinx)]dsinx =1/4ln(2+sinx)-1/4ln(2-sinx)+C
绛旓細鍏堝埄鐢ㄥ嶈鍏紡锛岀劧鍚庡埄鐢ㄥ垎閮绉垎娉曞強绗竴鎹㈠厓绉垎娉曪細鈭玿(sinx)^2 (cosx)^2dx =1/4 鈭玿(sin2x)^2dx =1/8 鈭玿(1-cos4x)dx =1/8 (鈭玿dx-鈭xcos4xdx)=1/16 x^2 - 1/64 xsin4x + 1/64 鈭玸in4xdx = 1/16 x^2 - 1/64 xsin4x - 1/256 cos4x + C ...
绛旓細鈭玸in²xdx= 1/2x -1/4sin2x + C銆侰涓绉垎甯告暟銆傝В绛旇繃绋嬪涓嬶細鏍规嵁涓夎鍏紡 sin²x = (1-cos2x) / 2锛屽彲寰楋細鈭 sin²x dx = (1/2) 鈭 (1-cos2x) dx = (1/2) ( x- (1/2)sin2x) + C = 1/2x -1/4sin2x + C ...
绛旓細鈭sin2xcos3xdx=(cosx)/2-(cos5x)/10+C銆傦紙C涓绉垎甯告暟锛夎В绛旇繃绋嬪涓嬶細鈭玸in2xcos3xdx =鈭1/2(sin(2x+3x)+sin(2x-3x))dx锛堢Н鍖栧拰宸級=1/2鈭玸in5xdx-1/2鈭玸inxdx =1/10鈭玸in5xd5x+1/2鈭玠cosx =(cosx)/2-(cos5x)/10+C ...
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绛旓細sin2xdx = 2sinxcosxdx = dsin^2x 鍥犱负(sin^2x)' = 2sinx*cosx.鎵浠(sin^2x) = 2sinxcosxdx
