請問∫(sinx)²dx怎麼解(附詳解謝謝)?
回答 (1)
Sol
A=∫sin^2 xdx
4A=4∫sin^2 xdx
=∫2-2cos(2x)dx
=2∫dx-∫2cos(2x)dx
=2∫dx-∫cos(2x)d(2x)
=2x-sin(2x)+c1
A=x/2-Sin(2x)/4+
收錄日期: 2021-04-30 22:31:45
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