求定积分,请大神帮帮忙,需要详细步骤!谢谢!拜托
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∫x²*cosx*dx
=x²*sinx - ∫2x*sinx*dx
=x²*sinx - 2∫x*sinx*dx
=x²*sinx - 2[x*(-cosx) - ∫(-cosx)*dx]
=x²*sinx - 2[-x*cosx + ∫cosx*dx]
=x²*sinx - 2[-x*cosx + sinx]
=[x²*sinx + 2x*cosx - 2sinx]|x=-π/2 → π/2
=[(π/2)²sin(π/2) + 2*(π/2)*cos(π/2) - 2sin(π/2)]
-[(-π/2)²sin(-π/2) + 2*(-π/2)*cos(-π/2) - 2sin(-π/2)]
=[π²/4 * 1 + π*0 - 2*1] - [π²/4 * (-1) + (-π)*0 - 2*(-1)]
=[π²/4 - 2] - [-π²/4 + 2]
=π²/2 - 4
=x²*sinx - ∫2x*sinx*dx
=x²*sinx - 2∫x*sinx*dx
=x²*sinx - 2[x*(-cosx) - ∫(-cosx)*dx]
=x²*sinx - 2[-x*cosx + ∫cosx*dx]
=x²*sinx - 2[-x*cosx + sinx]
=[x²*sinx + 2x*cosx - 2sinx]|x=-π/2 → π/2
=[(π/2)²sin(π/2) + 2*(π/2)*cos(π/2) - 2sin(π/2)]
-[(-π/2)²sin(-π/2) + 2*(-π/2)*cos(-π/2) - 2sin(-π/2)]
=[π²/4 * 1 + π*0 - 2*1] - [π²/4 * (-1) + (-π)*0 - 2*(-1)]
=[π²/4 - 2] - [-π²/4 + 2]
=π²/2 - 4
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