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求助一道不定积分？

2025-03-14 05:21:34

推荐回答（4个）

回答1：

利用第一类积分换元法，令x=sint，则dx=costdt，当x=1时，t=π/2，当x=0时，t=0，求解过程如下图所示：

回答2：

嗨，哥们，你的那个六次方放错位置了哦，那个应该是放到那个余弦上面去，而不是变量上面取得，而且那个应该不是六次方哦，你又怎么搞的搞得差很多那个嗯，应该是开放开出来之后应该是余弦的三次方财都要。

回答3：

利用降次公式cosx=(1-cos2x)/2
因为dx=cosudu
所以原式=∫(cosu)^4du
=∫(cos²u)²du
=1/4*∫(1-cos2u)²du
=1/4*∫(1-2cos2u+cos²2u)du
再用一次降次公司
=1/8*∫(2-4cos2u+1-cos4u)du=3u/8-1/4*sin2u-1/32*cos4u+C

回答4：

解如下图所示

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