∫cos3xcos2xdx怎样求它的定积分啊…… ∫cos3xcos2xdx的解题的详细解答
\u222bcos3xcos2xdx\u600e\u6837\u6c42\u5b83\u7684\u5b9a\u79ef\u5206\u554a \u4e0a\u9650\u4e3a1\u4e0b\u9650\u4e3a0\u7528\u79ef\u5316\u548c\u5dee\u516c\u5f0f\u53d8\u6362\u88ab\u79ef\u51fd\u6570
\u5148\u7531\u79ef\u5316\u548c\u5dee\u516c\u5f0f\u5f97\u5230
cos3xcos2x=0.5(cos5x+cosx)
\u6240\u4ee5
\u539f\u79ef\u5206
=\u222b0.5(cos5x+cosx) dx
=\u222b0.5cos5xdx +\u222b0.5cosxdx
=\u222b0.1cos5xd(5x) +\u222b0.5cosxdx
=0.1*sin5x +0.5*sinx +C\uff0cC\u4e3a\u5e38\u6570
积化和差:cosαcosβ=[cos（α+β）+cos（α-β）]/2
cos3xcos2x=[cos5x+cosx]/2
原式=1/10sin5x+1/2sinx+C
cos3xcos2x=1/2[cos5x+cosx]
所以,
∫cos3xcos2xdx=1/2*∫(cos5x+cosx)dx
=1/10*sin5x+1/2*sinx+C
C表示任意常数
∫cos3xcos2xdx
=(1/2)∫cos3xdsin2x
= (1/2)cos3xsin2x + (3/2)∫sin3xsin2xdx
=(1/2)cos3xsin2x - (3/4)∫sin3xdcos2x
=(1/2)cos3xsin2x - (3/4)sin3xcos2x +(9/4)∫cos3xcos2x dx
(-5/4)∫cos3xcos2xdx=(1/2)cos3xsin2x - (3/4)sin3xcos2x
∫cos3xcos2xdx=(-4/5)[(1/2)cos3xsin2x - (3/4)sin3xcos2x]+C
∫cos3xcos2xdx
=1/2∫(cos5x-cosx)dx
=1/10sin5x-1/2sinx+C
绛旓細鈭 cos3xcos2x dx =(1/2)鈭 (cos5x+cosx) dx =(1/10)sin5x + (1/2)sinx + C 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
绛旓細浠ヤ笂锛岃閲囩撼銆
绛旓細鍏堢敤绉寲鍜屽樊鍏紡锛屾妸cos3xcos2x鍖栨垚(cos5x+cosx)/2鐒跺悗鍘熷紡=(鈭玞os5xdx+鈭玞osxdx)/2=(sin5x)/10+(sinx)/2+C
绛旓細cos3xcos2x=1/2[cos5x+cosx]鎵浠,鈭玞os3xcos2xdx=1/2*鈭(cos5x+cosx)dx =1/10*sin5x+1/2*sinx+C C琛ㄧず浠绘剰甯告暟
绛旓細鈭玞os3xcos2x dx 灏辨槸绉垎鍜屽樊 =鈭1/2(cos(3x+2x)+cos(3x-2x))dx =1/2鈭玞os5xdx +1/2鈭玞osxdx =1/10鈭玞os5xd5x+1/2鈭玠sinx =sin5x/10+sinx/2+C
绛旓細鈭玞os3xcos2x dx 灏辨槸绉垎鍜屽樊 =鈭1/2(cos(3x+2x)+cos(3x-2x))dx =1/2鈭玞os5xdx +1/2鈭玞osxdx =1/10鈭玞os5xd5x+1/2鈭玠sinx =sin5x/10+sinx/2+C
绛旓細鍏堢敤绉寲鍜屽樊鍏紡 cos3x.cos2x=(1/2)(cos5x+cosx)鈭(cos3x.cos2x)dx =1/2(鈭玞os5x dx +鈭玞osx dx)=1/2(1/5 sin5x+sinx +C)=1/10 sin5x +1/2 sinx +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
绛旓細鈭(cos3xcos2x)dx =(1/2)鈭(cos3xcos2x+sin3xsin2x)+(cos3xcos2x-sin3xsin2x)dx =(1/2)鈭(cosx+cos5x)dx =(sinx)/2+(sin5x)/10+C 绫讳技鈭(cosaxcosbx)dx銆佲埆(sinaxcosbx)dx銆佲埆(sinaxsinbx)dx 閮藉彲浠ヨ繖鏍峰仛
