求不定积分∫sin2xcosxdx 求不定积分1/sin2xcosxdx
\u6c42\u4e0d\u5b9a\u79ef\u5206\u222b(sin2x/cosx)dx \u8c22\u8c22\uff01\u222b(sin2x/cosx)dx
=\u222b(2sinxcosx/cosx)dx
=\u222b2sinxdx
=-2cosx+C
\u222bdx/(sin2xcosx)
=(1/2)\u222bdx/[(sinx)(cosx)^2]
=(1/2)\u222b cscx (secx)^2dx
=(1/2)\u222b cscx dtanx
=(1/2)cscx. tanx +(1/2)\u222b tanx(cscxcotx) dx
=(1/2)cscx. tanx +(1/2)\u222b cscx dx
=(1/2)secx +(1/2)ln|cscx-cotx| + C
解：原式=2∫sinxcos²xdx=-2∫cos²xd(cosx)=-2[(1/3)cos³x]+C=-(2/3)cos³x+C
4。求定积分【0，1】∫{[arcsin(√x)]/√[x(1-x)]}dx
解：令arcsin(√x)=u，则√x=sinu，x=sin²u，dx=2sinucosudu；x=0时u=0；x=1时u=π/2；
故原式=【0，π/2】∫2usinucosudu/√[sin²u(1-sin²u)]=【0，π/2】∫2usinucosudu/(sinucosu)=【0，π/2】2∫udu=2(u²/2)【0，π/2】=π²/4.
因为sin2x = 2sinxcosx;
∫sin2xcosxdx = ∫2sinxcosxcosxdx = -2∫cosx^2dcosx = -2/3∫cosx^3
∫sin2xcosxdx
=∫2cosxsinxcosxdx
=－2∫cosxcosxdcosx
=－(2/3)(cosx)^3十C
绛旓細3銆姹備笉瀹氱Н鍒嗏埆sin2xcosxdx 瑙ｏ細鍘熷紡=2鈭玸inxcos²xdx=-2鈭玞os²xd(cosx)=-2[(1/3)cos³x]+C=-(2/3)cos³x+C 4銆傛眰瀹氱Н鍒嗐0锛1銆戔埆{[arcsin(鈭歺)]/鈭歔x(1-x)]}dx 瑙ｏ細浠rcsin(鈭歺)=u锛屽垯鈭歺=sinu锛寈=sin²u锛宒x=2sinucosudu锛泋=0鏃秛...
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
绛旓細鏍瑰彿涓(cos2x)鐨涓嶅畾绉垎姹傦細cos2x鐨勪笉瀹氱Н鍒嗘槸(1/2)sin2x+C銆傗埆cos2xdx =(1/2)鈭玞os2xd2x =(1/2)sin2x+C 鎵浠os2x鐨勪笉瀹氱Н鍒嗘槸(1/2)sin2x+C銆備笉鍙Н鍑芥暟锛氳櫧鐒跺緢澶氬嚱鏁伴兘鍙氳繃濡備笂鐨勫悇绉嶆墜娈佃绠楀叾涓嶅畾绉垎锛屼絾杩欏苟涓嶆剰鍛崇潃鎵鏈夌殑鍑芥暟鐨鍘熷嚱鏁閮藉彲浠ヨ〃绀烘垚鍒濈瓑鍑芥暟鐨勬湁闄愭澶嶅悎锛...
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绛旓細=(1/2)鈭(1-cos2x)dx =(1/2)x-(1/4)sin2x+C(C涓哄父鏁)涓嶅畾绉垎鐨勫叕寮忥細1銆佲埆adx=ax+C锛宎鍜孋閮芥槸甯告暟 2銆佲埆x^adx=[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/xdx=ln|x|+C 4銆佲埆a^xdx=(1/lna)a^x+C锛屽叾涓璦>0涓攁鈮1 5銆佲埆e^xdx=e^x+C 6銆鈭玞osxdx...
绛旓細鍥炵瓟锛氣埆cos(x)/(sin(x)^2)dx =鈭1/(sin(x)^2)dsin(x) =-1/sin(x)+C
绛旓細1,鈭玸in2xdx =鈭2sinxcosxdx=鈭2sinxd(sinx)=(sinx)^2+C 2,鈭玿cosxdx=xsinx-鈭玸inxdx=xsinx+cosx+c(鍒嗛儴绉垎娉)
绛旓細sinxcosx鐨涓嶅畾绉垎鏄細sinxcosx =1/2sin2x =1/4鈭玿sin2xdx =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...
绛旓細鈭玸in2xdx=2鈭玸inxcosxdx=2鈭玸inxdsinx 姝ｇ‘ 涔嬪悗浠inx =U 鍒欎负2鈭玌 dU =U*2 鍒欑瓑浜庯紙sinx锛*2
绛旓細瑙ｏ細鈭2sinxcos²xdx=-2鈭玞os²xd(cosx)=-2/3cos³x+c
