求COS3X·COS2X的不定积分？ 计算不定积分：积分号cos3xcos2xdx
cosxcos2xcos3x\u7684\u4e0d\u5b9a\u79ef\u5206cosxcos2xcos3x\u7684\u4e0d\u5b9a\u79ef\u5206\u4e3ax/4+1/8sin2x+1/16sin4x+1/24sin6x+C\u3002
\u89e3\uff1a\u222bcosxcos2xcos3xdx
=1/2\u222bcosx*(cos(3x+2x)+cos(3x-2x))dx
=1/2\u222bcosx*(cos5x+cosx)dx
=1/2\u222bcosxcos5xdx+1/2\u222b(cosx)^2dx
=1/4\u222b(cos(5x+x)+cos(5x-x))dx+1/4\u222b(1+cos2x)dx
=1/4\u222b1dx+1/4\u222bcos2xdx+1/4\u222bcos6xdx+1/4\u222bcos4xdx
=x/4+1/8sin2x+1/16sin4x+1/24sin6x+C
\u6269\u5c55\u8d44\u6599\uff1a
1\u3001\u4e09\u89d2\u51fd\u6570\u79ef\u5316\u548c\u5dee\u516c\u5f0f
\uff081\uff09cosAcosB=1/2*(cos(A+B)+cos(A-B))
\uff082\uff09sinAsinB=1/2*(cos(A-B)-cos(A+B))
\uff083\uff09cosAsinB=1/2*(sin(A+B)-sin(A-B))
\uff084\uff09sinAcosB=1/2*(sin(A+B)+sin(A-B))
2\u3001\u4e8c\u500d\u89d2\u516c\u5f0f
sin2A=2sinAcosA\u3001cos2A=(cosA)^2-(sinA)^2=2(cosA)^2-1=1-2(sinA)^2
3\u3001\u4e0d\u5b9a\u79ef\u5206\u516c\u5f0f
\u222b1dx=x+C\u3001\u222bcosxdx=sinx+C\u3001\u222be^xdx=e^x+C
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u4e0d\u5b9a\u79ef\u5206
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u5229\u7528\u79ef\u5316\u548c\u5dee\u516c\u5f0f\u516c\u5f0f
\u5c06\u79ef\u5206\u51fd\u6570\u5316\u4e3a\u548c\u7684\u5f62\u5f0f
\u518d\u5206\u522b\u51d1\u5fae\u5206
\u8fc7\u7a0b\u5982\u4e0b\u56fe\uff1a
=∫1/2(cos(3x+2x)+cos(3x-2x))dx
=1/2∫cos5xdx +1/2∫cosxdx
=1/10∫cos5xd5x+1/2∫dsinx
=sin5x/10+sinx/2+C
利用积化和差公式，cosacosb=1/2[cos(a+b)–cos(a–b)] 可以算的积分为1/10×sin5x+sinx/2+c
利用积化和差公式，cosacosb=1/2[cos(a+b)+cos(a–b)] ,化简后求积分
COS3X·COS2X=1/2[cos(3x+2x)+cos(3x-2x)]=1/2cos5x+1/2cosx
积分=1/10sin5x+1/2sinx+C
绛旓細绉寲鍜屽樊鍏紡锛堜腑瀛︾煡璇嗭級cosxcosy=(1/2)[cos(x+y)+cos(x-y)]鍥犳锛cos3xcos2x=(1/2)(cos5x+cosx)銆愭暟瀛︿箣缇庛戝洟闃熶负鎮ㄨВ绛旓紝鑻ユ湁涓嶆噦璇疯拷闂紝濡傛灉瑙ｅ喅闂璇风偣涓嬮潰鐨勨滈変负婊℃剰绛旀鈥濄
绛旓細涓嶇敤浠栦滑閭ｄ箞楹荤儲!绉寲鍜屽樊:cos伪cos尾=[cos锛埼+尾锛+cos锛埼-尾锛塢/2 cos3xcos2x=[cos5x+cosx]/2 鍘熷紡=1/10sin5x+1/2sinx+C
绛旓細cosxcos2x=(1/2)锛坈osx+cos3x)鎵浠涓嶅畾绉垎鏄(1/2)sinx+(1/6)sin3x+C cos3x路sinx=(1/2)(sin4x-sin2x)鎵浠ヤ笉瀹氱Н鍒嗘槸(1/4)cos2x-(1/8)cos4x+C
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绛旓細鍏蜂綋鏉ヨ锛屽彲浠ヤ娇鐢ㄤ笁瑙掑嚱鏁扮殑鍜屽樊鍏紡灏cos3x琛ㄧず涓篶os(2x + x)锛岀劧鍚庢牴鎹甤os(a + b)鐨勫叕寮忓睍寮寰楀埌cos2xcosx - sin2xsinx銆傛帴鐫锛屽皢cos2x鍜宻in2x鍒嗗埆鐢2cos²x - 1鍜2sinxcosx鏇挎崲锛屽緱鍒(2cos²x - 1)cosx - 2(1 - cos²x)cosx銆傛渶鍚庡寲绠寰楀埌4cos³x - 3...
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绛旓細浣欏鸡鍑芥暟鐨勪笁鍊嶈鍏紡鏄細鎺ㄥ杩囩▼濡備笅锛cos3x=cos(x+2x)=cosx*cos2x-sinx*sin2x =cosx*(2cosx^2-1)-sinx*(2sinxcosx)=2cosx^3-cosx-2cosx(1-cosx^2)=4cosx^3-3cosx
绛旓細杩欓亾棰樺苟涓嶆槸鐗瑰埆绠鍗曪紝鍏蜂綋杩囩▼濡備笅锛氫互涓婏紝璇烽噰绾炽
绛旓細杩欎釜鏄掓暟鐨勫洓鍒欒繍绠楀拰绗﹀悎鍑芥暟鐨勫鏁 (abc)'=a'bc+ab'c+abc'(cosxcos2xcos3x)'=-sinxcos2xcos3x-2cosxsin2xcos3x-3cosxcox2xsin3x 鏈鍚庣畻鍑虹殑缁撴灉鏄14鍚
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