∫cosx cos2xcos3xdx
cosxcos2xcos3x的不定积分为x/4+1/8sin2x+1/16sin4x+1/24sin6x+C。
解:∫cosxcos2xcos3xdx
=1/2∫cosx*(cos(3x+2x)+cos(3x-2x))dx
=1/2∫cosx*(cos5x+cosx)dx
=1/2∫cosxcos5xdx+1/2∫(cosx)^2dx
=1/4∫(cos(5x+x)+cos(5x-x))dx+1/4∫(1+cos2x)dx
=1/4∫1dx+1/4∫cos2xdx+1/4∫cos6xdx+1/4∫cos4xdx
=x/4+1/8sin2x+1/16sin4x+1/24sin6x+C
扩展资料:
1、三角函数积化和差公式
(1)cosAcosB=1/2*(cos(A+B)+cos(A-B))
(2)sinAsinB=1/2*(cos(A-B)-cos(A+B))
(3)cosAsinB=1/2*(sin(A+B)-sin(A-B))
(4)sinAcosB=1/2*(sin(A+B)+sin(A-B))
2、二倍角公式
sin2A=2sinAcosA、cos2A=(cosA)^2-(sinA)^2=2(cosA)^2-1=1-2(sinA)^2
3、不定积分公式
∫1dx=x+C、∫cosxdx=sinx+C、∫e^xdx=e^x+C
参考资料来源:百度百科-不定积分
参考资料来源:百度百科-三角函数公式
cosxcos2xcos3x
= cos2x * cosxcos3x
= cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)]
= cos2x * (1/2)[cos4x + cos2x]
= (1/2)cos2xcos4x + (1/2)cos²2x
= (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x)
= (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4x
∫ cosxcos2xcos3x dx
= (1/4)∫ dx + (1/4)∫ cos2x dx + (1/4)∫ cos4x dx + (1/4)∫ cos6x dx
= x/4 + (1/8)sin2x + (1/16)sin4x + (1/24)sin6x + C
积化和差:
2cosxcos3x=cos4x+cos2x
4cosxcos2xcos3x=2cos2x[cos4x+cos2x]
=cos6x+cos2x+cos4x+1
【2】
可设原式=y
4y=∫[cos6x+cos4x+cos2x+1]dx
=(1/6)∫cos6xd(6x)+(1/4)∫cos4xd(4x)+(1/2)∫cos2xd(2x)+∫dx
=[(sin6x)/6]+[(sin4x)/4]+[(sin2x)/2]+x+C
∴原式=【。。。。。。]/4
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