不定积分cos^6(3x)dx
2个回答
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=∫[(1+cos6x)/2]³dx
=1/8∫(1+3cos6x+3cos²6x+cos³6x)dx
=1/8[∫dx+3∫cos6xdx+∫3cos²6xdx+∫cos³6xdx]
=1/8[x+1/2∫cos6xd6x+1/4∫(1+cos12x)/2d12x+1/6∫(1-sin²6x)dsin6x]
=1/8[x+1/2*sin6x+1/4*(6x+sin12x/2)+1/6*(sin6x-sin³6x/3)]+C
=5x/16+11/48*sin6x+1/64*sin12x-1/18*sin³6x+C
=1/8∫(1+3cos6x+3cos²6x+cos³6x)dx
=1/8[∫dx+3∫cos6xdx+∫3cos²6xdx+∫cos³6xdx]
=1/8[x+1/2∫cos6xd6x+1/4∫(1+cos12x)/2d12x+1/6∫(1-sin²6x)dsin6x]
=1/8[x+1/2*sin6x+1/4*(6x+sin12x/2)+1/6*(sin6x-sin³6x/3)]+C
=5x/16+11/48*sin6x+1/64*sin12x-1/18*sin³6x+C
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