| ■13820 / inTopicNo.2) |
Re[1]: 定積分
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□投稿者/ laki 一般人(5回)-(2006/06/22(Thu) 20:12:58)
 | ①∫x^1/2dx=[1/(2x^1/2)]=(1-√2)/4
②∫e^xdx=[e^x]=e-1/e
③∫1/xdx=[logx]=log3
①∫cos^2xdx=∫1/2(1+cos2x)=1/2x+1/4sinx
②∫sin^3xdx=∫(3/4sinx-1/4sin3x)dx=-3/4cosx+1/12cos3x
③∫sin3xcosxdx=∫(3sinxcosx-4sin^3xcosx)dx
=∫[{(3/2(sinx)^2}'-(sin^4x)']dx
=3/2sin^2x-sin^4x
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