設(shè)x=sint,dx=costdt,t∈[0,2π]
原式=∫(sint)^2(cost)^2dt
=(1/4)∫(sin2t)^2dt
=(1/8))∫(1-cos4t)dt
=(t/8)-(1/32)∫cos4td(4t)
=(t/8)-(1/32)sin4t·4t
把x帶回去就可以了
∫x².√（1-x²）dx,
∫x².√（1-x²）dx,
求不定積分
求不定積分
其他人氣：929 ℃時間：2020-03-26 07:16:03
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