后面就是基础部分了,不用详述了吧.
∫1/2(sin8x-sin2x)dx=-1/16cos8x+1/4cos2x+C
求不定积分∫sin3xcos5xdx 请高手详细阐述下解题思路,谢谢!
sin3xcos5x=1\/2(sin8x-sin2x) 积化和差 后面就是基础部分了,不用详述了吧.∫1\/2(sin8x-sin2x)dx=-1\/16cos8x+1\/4cos2x+C
∫sin3xcos5x dx求不定积分 没学积化和差
解答过程如下：∫sin3x*cos5xdx =1\/2*∫[sin(5x+3x) - sin(5x -3x)]dx =1\/2*∫[sin8x - sin2x]dx =1\/2*∫sin8x*dx - 1\/2*∫sin2xdx =1\/2*1\/8*∫sin8x*d(8x) - 1\/2*1\/2*∫sin2xd(2x)=-1\/16*cos8x + 1\/4*cos2x + C ...
求不定积分∫sin3xcos5xdx
使用积化和差公式：∫sin3x*cos5x*dx=1\/2*∫[sin(5x+3x) - sin(5x -3x)]*dx=1\/2*∫[sin8x - sin2x]*dx=1\/2*∫sin8x*dx - 1\/2*∫sin2x*dx=1\/2*1\/8*∫sin8x*d(8x) - 1\/2*1\/2*∫sin2x*d(2x)=-1\/16*cos8x + 1\/4*cos2x + ...
求不定积分 ∫sin^3x*cos^5x 乘dx
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
求sin3x*cos5x的不定积分
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大一高数～求不定积分∫sin3xsin5xdx
须知公式：sinAsinB = (1\/2)[cos(A-B)-cos(A+B]∫sin(3x)sin(5x) dx= (1\/2)∫[cos(-2x)-cos(8x)] dx= (1\/2)∫cos2x dx - (1\/2)∫cos8x dx= (1\/4)∫cos2x d(2x) - (1\/16)∫cos8x d(8x)= (1\/4)sin2x - (1\/16)sin8x + C ...
用凑微分法求∫sin3xcos5xdx,这里的3X和5X不是乘积,是sinx的3次方乘以...
∫(sinx)^3.(cosx)^5dx = ∫-(1-(cosx)^2) (cosx)^5 dcosx = (cosx)^8\/8 - (cosx)^6\/6 +C
高数中的不定积分
cos8x=cos(3x+5x)=cos3xcos5x-sin3xsin5x cos2x=cos(-2x)=cos(3x-5x)=cos3xcos5x+sin3xsin5x 故sin3xsin5x=1\/2*(cos2x-cos8x)∫Sin3xSin5xdx=∫1\/2*(cos2x-cos8x)dx=1\/4*sin2x-1\/16*sin8x+C
不定积分求解谢谢过程
sin(A-B)=sinA.cosB-cosA.sinB (2)(1)+(2)sinA.cosB =(1\/2)[sin(A+B)+sin(A-B)]A=2x, B=3x sin2x.cos3x =(1\/2)[sin5x-sinx]∫sin2x.cos3x dx =∫(1\/2)[sin5x-sinx] dx =(1\/2)[ -(1\/5)cos5x + cosx ] +C =(1\/2)cosx -(1\/10)cos5x + C ...
不定积分!!
方法如下，请作参考：36：40：
