(sinx)^6dx=?
1个回答
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先求不定积分
∫(sinx)^6dx
=∫[(1-cos2x)/2]³dx
=1/8∫(1-3cos2x+3cos²2x-cos³2x)dx
=1/8∫[1-3cos2x+3(1+cos4x)/2-cos³2x]dx
=1/8∫(5/2-3cos2x+3/2cos4x-cos³2x)dx
=1/8[∫5/2dx-3∫cos2xdx+3/2∫cos4xdx-∫cos³2xdx]
=1/8[5x/2-3/2sin2x+3/8sin4x-1/2∫(1-sin²2x)d(sin2x)]
=1/8[5x/2-3/2sin2x+3/8sin4x-1/2(sin2x-1/3sin³2x)]+C
=1/8[5x/2-2sin2x+3/8sin4x+1/6sin³2x]+C
然后求微分
那么(sinx)^6dx=1/8d (5x/2-2sin2x+3/8sin4x+1/6sin³2x)
∫(sinx)^6dx
=∫[(1-cos2x)/2]³dx
=1/8∫(1-3cos2x+3cos²2x-cos³2x)dx
=1/8∫[1-3cos2x+3(1+cos4x)/2-cos³2x]dx
=1/8∫(5/2-3cos2x+3/2cos4x-cos³2x)dx
=1/8[∫5/2dx-3∫cos2xdx+3/2∫cos4xdx-∫cos³2xdx]
=1/8[5x/2-3/2sin2x+3/8sin4x-1/2∫(1-sin²2x)d(sin2x)]
=1/8[5x/2-3/2sin2x+3/8sin4x-1/2(sin2x-1/3sin³2x)]+C
=1/8[5x/2-2sin2x+3/8sin4x+1/6sin³2x]+C
然后求微分
那么(sinx)^6dx=1/8d (5x/2-2sin2x+3/8sin4x+1/6sin³2x)
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