1.∫ [sin(2x)]^2 dx 2.∫(1-tanx)/(1+tanx) dx
人气:282 ℃ 时间:2020-05-14 12:29:01
解答
1.∫ [sin(2x)]^2 dx=∫(1-cos4x)/2 dx
=(1/2)x-(1/8)∫cos4xd(4x)
=(1/2)x-(1/8)sin4x+C
2.∫(1-tanx)/(1+tanx) dx
=∫(cosx-sinx)/(cosx+sinx) dx
=∫1/(cosx+sinx)d(cosx+sinx)
=ln|cosx+sinx|+C
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