粉红内衣
春芽
共回答了21个问题采纳率：81% 举报
∵∫x²sinxdx=(-x²cosx+2xsinx+2cosx)│ (应用分部积分法)
=π²-2-2
=π²-4
∫sin³xdx=∫(1-cos²x)sinxdx
=∫(cos²x-1)d(cosx)
=(cos³x/3-cosx)│
=-1/3+1-1/3+1
=4/3
∴∫∫(x²-y²)dxdy=∫dx∫(x²-y²)dy
=∫(x²sinx-sin³x/3)dx
=∫x²sinxdx-(1/3)∫sin³xdx
=(π²-4)-(1/3)(4/3)
=π²-40/9.
1年前
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