求(x^3+1)cos^2x 的不定积分
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求(x^3+1)cos^2x 的不定积分
![求(x^3+1)cos^2x 的不定积分](/uploads/image/z/16136355-3-5.jpg?t=%E6%B1%82%28x%5E3%2B1%29cos%5E2x+%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86)
∫(x^3+1)(cosx)^2dx
=∫(x^3+1)[(1+cos2x)/2]dx
=(1/2)∫(x^3+1)dx+(1/2)∫cos2xdx+(1/2)∫x^3cos2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x *x^3 -(1/4)3x^2sin2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x*x^3 +(3/8)cos2x*x^2-(3/4)∫xcos2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x*x^3+(3/8)cos2x*x^2-(3/8)sin2x*x-(3/16)cos2x+C
=∫(x^3+1)[(1+cos2x)/2]dx
=(1/2)∫(x^3+1)dx+(1/2)∫cos2xdx+(1/2)∫x^3cos2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x *x^3 -(1/4)3x^2sin2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x*x^3 +(3/8)cos2x*x^2-(3/4)∫xcos2xdx
=(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x*x^3+(3/8)cos2x*x^2-(3/8)sin2x*x-(3/16)cos2x+C