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2Cosx2的不定积分

答案为 1/2x+1/4sin2x+C.解题过程:解:原式=1/2∫(1+cos2x)dx=1/2∫1dx+1/2∫cos2xdx=1/2x+1/4∫cos2xdx=1/2x+1/4sin2x+C 如果看不懂文字的格式,可以看图片.拓展资料 不定积分的简介:在微积分中,一个函数f 的不定积分,或原函数,或反导数,是一个导数等于f 的函数 F ,即F ′ = f.不定积分和定积分间的关系由微积分基本定理确定.其中F是f的不定积分.

sin2X

先用积化和差公式,再求积分即可cosxcosy=(1/2)[cos(x+y)+cos(x-y)]∫[cosx*cos(x/2)]dx=1/2*∫[cos(x+x/2)+cos(x-x/2)]dx=1/2*∫[cos(3x/2)+cos(x/2)]dx=1/2*∫cos(3x/2)dx+1/2*∫cos(x/2)dxd(3x/2)=(3/2)dx,d(x/2)=(1/2)dx=1/2*2/3*∫cos(3x/2)d(3x/2)+1/2*2*∫cos(x/2)d(x/2)=1/3*sin(3x/2)+sin(x/2)+C=(1/3)sin(3x/2)+sin(x/2)+C

∫ cosx dx=(1/2)∫ (1+cos2x) dx=(1/2)x + (1/4)sin2x + c【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”.

没可能降幂吧?醒目点的都知道要用积化和差公式啦! cosxcosy=(1/2)[cos(x+y)+cos(x-y)] ∫[cosx*cos(x/2)]dx =1/2*∫[cos(x+x/2)+cos(x-x/2)]dx =1/2*∫[cos(3x/2)+cos(x/2)]dx =1/2*∫cos(3x/2)dx+1/2*∫cos(x/2)dx d(3x/2)=(3/2)dx,d(x/2)=(1/2)dx =1/2*2/3*∫cos(3x/2)d(3x/2)+1/2*2*∫cos(x/2)d(x/2) =1/3*sin(3x/2)+sin(x/2)+c =(1/3)sin(3x/2)+sin(x/2)+c

(cosx)^2(sinx)^3 的不定积分∫(cosx)^2(sinx)^3 dx=-∫(cosx)^2(sinx)^2 dcosx=-∫(cosx)^2(1-(cosx)^2) dcosx= -[ (cosx)^3/3- (cosx)^5/5] + c

∫2xcosx∧2dx=∫x(cos2x+1)dx=x^2/2+(1/2)∫xdsin2x=x^2/2+(1/2)xdsin2x-(1/2)∫sin2xdx=x^2/2+(1/2)xdsin2x+(1/4)∫dcos2x=x^2/2+(1/2)xdsin2x+(1/4)dcos2x+c

∫cos2xcosxdx =∫[1-2(sinx)^2]d(sinx) =∫d(sinx)-2∫(sinx)^2d(sinx) =sinx-(2/3)(sinx)^3+C =(1/3)sinx[3-2(sinx)^2]+C =(1/3)sinx(2+cos2x)+C =(2/3)sinx+(1/3)sinxcos2x+C.

∫xcos(x/2)dx=2∫xcos(x/2)d(x/2)=2∫ xdsin(x/2)=2xsin(x/2)-2∫ sin(x/2)dx=2xsin(x/2)+4cos(x/2)+C

你好!∫ 2cosx dx = 2sinx +C若是定积分请给出积分区间

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