∫cos(1-3x)dx

∫cos((x/3)-1)dx=?3sin((x/3)-1)3sin((x/3)-1)这是凑微分法。把后面的dx凑成d（x/3-1）∫cos((x/3)-1)dx=3*∫cos((x/3)-1)d((x/3)-1)=3sin((x/3)-1

∫cos(3x-1)dxScos(3x-1)dx=1/3*Scos(3x-1)d（3x-1）=1/3*sin(3x-1)(1/3)*sin(3x-1)+a

/(cosx+1)dx也可以考虑,分子分母同时乘以1-cosx,被积函数化为：(1-cosx)/sin²xI=∫(1-cosx)/sin²xdx=∫[csc²x-cscxcotx]dx=-cotx+cscx+C

不定积分∫x*cos(x/3)*dx分部积分法∫udv=uv-∫vduu=3x,v=sin(x/3)结果是3xsin(x/3)+9cos(x/3)分部积分，，cos拿到d后面，，答案：3xsin(x/3)+9cos(x/3)+C。

∫cos(x-1)dx、∫x^3e^x^2dx怎么解∫cos(x-1)dx=∫cos(x-1)d(x-1)=sin(x-1)+C∫x³e^(x²)dx-->令u=x²,du=2xdx-->=∫uxe^u·du/

∫cos^3xdx∫tan^3xsecxdx∫(cosx)^3dx=∫(1-sinx^2)dsinx=sinx-(1/3)(sinx)^3+C∫tanx^3secxdx=∫tanx^2dsecx=∫(secx^2-1)dsecx=(1/3)

∫(1+cos^3x)dx的不定积分∫(1+cosx^3)dx=x+∫cosx^3dx=x+∫(1-sinx^2)dsinx=x+sinx-(sinx)^3/3+C=∫dx+∫cos³xdx=x+∫cos²xcosxdx

求∫cos（3x+1）dx.∫cos(3x+1)dx=(1/3)∫cos(3x+1)d3x=(1/3)∫cos(3x+1)d(3x+1)=(1/3)∫d(sin(3x+1))=(1/3)sin(3x+1)+C

求解不定积分∫cos(3x+1)dx∫cos(3x+1)dx=(1/3)∫cos(3x+1)d3x=(1/3)∫cos(3x+1)d(3x+1)=(1/3)∫d(sin(3x+1))=(1/3)sin(3x+1)+C解∫cos(3x+1)d

∫1/(1+cos^2(x))dx∫dx/{1+[cos(x)]^2}=∫[sec(x)]^2dx/{1+[sec(x)]^2}=∫[sec(x)]^2dx/{2+[tan(x)]^2}=∫2^(-1/2)d[tan(x)/2^(1/2)]

∫1/1-cosxdx原式=∫dx/(2sin²(x/2))=1/2∫csc²(x/2)dx=∫csc²(x/2)d(x/2)=-cot(x/2)+C(C是积分常数).

∫(sinx/cos^3x)dx∫(sinx/cos^3x)dx=-∫(dcosx/cos^3x)=1/2cos^2x

∫(1-sin/x+cos)dx不定积分可用凑微分法如图积分.经济数学团队帮你解答,请及时采纳.

∫(1+sinx)/cos^2xdx∫(1+sinx)/(cosx)^2dx=∫[(secx)^2+tanxsecx]dx=tanx+secx+C

∫(1-cos^(2)2x)dx=∫(1-cos4x)/2dx=∫1/2dx-∫cos4x/8d4x=0.5x-1/8*sin4x+C(C为任意常数)

∫sin^(3)xcos^(3)xdx如何求出答案为1/5cos^(5)x-1/3cos^(3)x+C计算过程是怎样的?解法一：第一换元法（凑微分法）∫sin³xcos³xdx=∫sin²xcos³x

求不定积分!∫cos^(-1)xdx∫cos√xdx∫cos^(-1)xdx=∫secxdx=ln|secx+tanx|+c∫cos√xdx令√x=t,x=t²,dx=2tdt原式=2∫tcostdt=2∫tdsint=2tsin

∫cos²xdx∫cos²xdx=(1/2)∫(1+cos2x)dx=……

∫cos³xdx∫cos³xdx=∫cosx*（1-2sin²x）dx=∫cosxdx-2∫cosxsin²xdx=sinx-2∫sin²xd（sinx）=sinx-2/3sin³

求不定积分1)∫3^(-x)*(2*3^x-3*2^x)dx2)∫cos^2(x/2)dx1）2x+[3(2/3)^x]/(ln3-ln2)+c2)1/2(x+sinx)+c1)∫3^(-x)*(2*3^x-3*2^x)dx=∫(2-3*(