∫x^2sin(x^3)dx换元

∫4/(1-2x)^2dx∫1/(3x+5)dx利用换元积分法求不定积分~∫4/(1-2x)^2dx∫1/(3x+5)dx利用换元积分法求不定积分~∫4/(1-2x)^2dx∫1/(3x+5)dx利用

换元积分题目∫[(dx)/(2-3x)^(1/3)]求以换元积分做法计算换元积分题目∫[(dx)/(2-3x)^(1/3)]求以换元积分做法计算换元积分题目∫[(dx)/(2-3x)^(1/3)]求以

不定积分∫1/（x+a）dx,∫1/根号（2-5x）dx,∫1/根号2-3x^2dx,用第一类换元积分法做,不定积分∫1/（x+a）dx,∫1/根号（2-5x）dx,∫1/根号2-3x^2dx,用第一

换元积分法∫x/(x^2-x-2)dx换元积分法∫x/(x^2-x-2)dx换元积分法∫x/(x^2-x-2)dx∫xdx/(x^2-x-2)=(1/3)∫[(x-2)+2*(x+1)]dx/[(x-

利用换元积分法求∫1/{√[16+6x-(x^2)]}dx利用换元积分法求∫1/{√[16+6x-(x^2)]}dx利用换元积分法求∫1/{√[16+6x-(x^2)]}dx原式=∫dx/√[25-(

∫x/sin^2(x)dx∫x/sin^2(x)dx∫x/sin^2(x)dx原式=-∫xd(cotx)=-xcotx+∫cotxdx=-xcotx+∫cosx/sinx*dx=-xcotx+∫d(s

∫sin(x)cos^2(x)dx∫sin(x)cos^2(x)dx∫sin(x)cos^2(x)dx原式=-∫cos²xdcosx=-cos³x/3+C

∫（3x-2）^10dx∫根号下（2+3x）dx利用换元积分法求不定积分,∫（3x-2）^10dx∫根号下（2+3x）dx利用换元积分法求不定积分,∫（3x-2）^10dx∫根号下（2+3x）dx利用

换元积分法这是我的做法∵d(sin(x/2))=cos(x/2)/2dx∴cos(x/2)dx=2d(sin(x/2))∴原式=∫2cosxd(sin(x/2))=2∫1-2sin^2(x/2)d((

∫(cos^3x/sin^2x)dx∫(cos^3x/sin^2x)dx∫(cos^3x/sin^2x)dx∫(cos^3x/sin^2x)dx=∫[(1-(sinx)^2]/(sinx)^2dsin

∫(x^3)*(sinx^2)dx求不定积分∫(x^3)*(sinx^2)dx求不定积分∫(x^3)*(sinx^2)dx求不定积分解∫x³sinx²dx=1/2∫x²s

∫[cos^3(x)]/[sin^2(x)]dx∫[cos^3(x)]/[sin^2(x)]dx∫[cos^3(x)]/[sin^2(x)]dx∫[cos^3(x)]/[sin^2(x)]dx=积分:

∫sin^3(x)cos^2(x)dx=∫sin^3(x)cos^2(x)dx=∫sin^3(x)cos^2(x)dx=把一个sin(x)拿出来∫sin^3(x)cos^2(x)dx=-∫sin^2(

求积分：∫sin^2(x)/cos^3(x)dx求积分：∫sin^2(x)/cos^3(x)dx求积分：∫sin^2(x)/cos^3(x)dx原式等于：∫[1-cos^2(x)]/cos^3(x)d

高数∫(x^2)(sin[x^3])dx高数∫(x^2)(sin[x^3])dx高数∫(x^2)(sin[x^3])dx望采纳=-cos[x^3]/3+C1/3（-cosx3)+c

第二换元积分法求dx/√(x^2+1))^3的积分第二换元积分法求dx/√(x^2+1))^3的积分第二换元积分法求dx/√(x^2+1))^3的积分第二换元将x换成tanθ原积分=∫cos^3θdt

换元积分法∫（1/1+√2x)dx令t=√2xdx=tdt请问dx=tdt是怎么算的?换元积分法∫（1/1+√2x)dx令t=√2xdx=tdt请问dx=tdt是怎么算的?换元积分法∫（1/1+√2x

∫sin^2x/(1+sin^2x)dx求解,∫sin^2x/(1+sin^2x)dx求解,∫sin^2x/(1+sin^2x)dx求解,原式=∫[1-1/(2sin²x+cos²

求∫√(sin^3x-sin^5x)dx求∫√(sin^3x-sin^5x)dx求∫√(sin^3x-sin^5x)dx我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积

∫sin(6x)sin(2x)dx=?∫sin(6x)sin(2x)dx=?∫sin(6x)sin(2x)dx=?利用积化和差公式cos(6x+2x)=cos8x=cos6xcos2x-sin6xsi