cos+5x的不定积分

项洁维3057不定积分:sinx^3cosx^2 -
尤国庭19889575852 ______ ∫sinx^3cosx^2dx=-∫sin^2xcos^2xdcosx=-∫(1-cos^2x)cos^2xdcosx=-∫(-cos^4x+cos^2x)dcosx=(1/5)cos^5x-(1/3)cos^3x+C

项洁维3057求不定积分 ∫sin^3x*cos^5x 乘dx -
尤国庭19889575852 ______[答案] 答:∫[(sinx)^3(cosx)^5]dx = - ∫[(sinx)^2(cosx)^5]d(cosx) = - ∫[(cosx)^5-(cosx)^7]d(cosx) =(1/8)(cosx)^8-(1/6)(cosx)^6+C

项洁维3057各位学长帮帮忙高数不定积分∫cos(5X+2)dx=∫1/5cos(5X+2)*(5X+2)＂dx 注明:(＂)是导数那个符号请问等号后面的那个1/5是怎么推出来的 谁能帮我解释下 -
尤国庭19889575852 ______[答案] 5x+2的导数是5,所以dx可以化成:dx=1/5*(5x+2)'dx=1/5d(5x+2),这是第一类换元法里面的一种定式,即 a≠0,∫f(ax+b)dx=1/a*∫f(ax+b)d(ax+b)=1/a*∫f(u)du,u=ax+b.只要知道了f(u)的原函数,即可得到原不定积分的结果

项洁维3057不定积分∫XCOS(5X^2+1)DX怎么求?万谢! -
尤国庭19889575852 ______[答案] ∫XCOS(5X^2+1)DX =1/2∫cos(5x^2+1)dx^2 =1/10∫cos(5x^2+1)d(5x^2+1) =1/10sin(5x^2+1)+C

项洁维3057求不定积分∫sin(3x)sin(5x)dx,求详细过程~ -
尤国庭19889575852 ______[答案] sin(3x)sin(5x) =sin(4x-x)sin(4x+x) =(sin4xcosx-cos4xsinx)*(sin4xcosx+cos4xsinx) = sin^4x-sin^x =(1-cos8x)/2-(1-cosx)/2 =(cos2x-cos8x)/2 所以∫sin(3x)sin(5x)dx=sin2x/4-sin8x/16+C

项洁维3057不定积分(sin5xcos3x) -
尤国庭19889575852 ______[答案] 先积化和差,再积分!sin5xcos3x=0.5sin(8x)+0.5sin(2x)∫sin5xcos3x dx=∫0.5sin(8x)+0.5sin(2x)dx=(1/16)∫sin(8x)d(8x) + (1/4)∫sin(2x)d(2x)=-(1/16)cos(8x) - (1/4)cos(2x) + C

项洁维3057cosx^6的不定积分 -
尤国庭19889575852 ______ 主要是在化简指数问题上,尽量把次方形式变为复角形式,例如cos(nx)和sin(nx)等,比较好积cos^6x = (cos²)³= [(1+cos2x)/2]³ = (1/8)(1+cos2x)³= (1/8)(1+3cos2x+3cos²2x+cos³2x)= (1/8)+(3/8)cos2x+(3/8)(1/2)(1+cos4x)+(1/8)(1/2)(1+cos4x...

项洁维3057积分cos(5x - 2)dx求解 -
尤国庭19889575852 ______[答案] ∫ cos(5x-2) dx =(1/5)∫ cos(5x-2) d(5x) =(1/5)∫ cos(5x-2) d(5x-2) =sin(5x-2)/5+C C为任意常数

项洁维3057关于不定积分的求1/cos∧5x的不定积分! -
尤国庭19889575852 ______[答案] K1=∫secxdx =∫secx(secx+tanx)/(secx+tanx)dx =∫d(secx+tanx)/(ecx+tanx) =ln|secx+tanx|+C k2=∫sec^5dx =∫sec^3xdtanx =... 即: k2=(1/4)tanxsec^3x+(3/4)∫sec^3xdx k3=∫sec^3xdx =∫secxdtanx =secxtanx-∫tanxdsecx =secxtanx-∫(1-cos^2x)dx/cos^3x =...

项洁维3057∫cos(5x - π/6)dx不定积分= -
尤国庭19889575852 ______[答案] ∫cos(5x-π/6)dx =1/5∫cos(5x-π/6)d(5x-π/6) =1/5*sin(5x-π/6)+C