一到微积分这题如何解,/>
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/08 14:10:53
一到微积分
![](http://img.wesiedu.com/upload/5/95/59577ee853569d694324625d4f34d9aa.jpg)
这题如何解,/>
![](http://img.wesiedu.com/upload/5/95/59577ee853569d694324625d4f34d9aa.jpg)
这题如何解,/>
![一到微积分这题如何解,/>](/uploads/image/z/19435254-6-4.jpg?t=%E4%B8%80%E5%88%B0%E5%BE%AE%E7%A7%AF%E5%88%86%E8%BF%99%E9%A2%98%E5%A6%82%E4%BD%95%E8%A7%A3%2C%2F%3E)
/>令√x=u,
则:x=u^2,dx=2udu.
∴∫[arcsin√x/√(1-x)]dx
=∫[arcsinu/√(1-u^2)]2udu
=-2∫arcsinu{-2u/[2√(1-u^2)]}du
=-2∫arcsinud[√(1-u^2)]
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)]d(arcsinu)
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)][1/√(1-u^2)]du
=-2[√(1-u^2)]arcsinu+2∫du
=2u-2[√(1-u^2)]arcsinu+C
=2√x-2[√(1-x)]arcsin√x+C
则:x=u^2,dx=2udu.
∴∫[arcsin√x/√(1-x)]dx
=∫[arcsinu/√(1-u^2)]2udu
=-2∫arcsinu{-2u/[2√(1-u^2)]}du
=-2∫arcsinud[√(1-u^2)]
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)]d(arcsinu)
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)][1/√(1-u^2)]du
=-2[√(1-u^2)]arcsinu+2∫du
=2u-2[√(1-u^2)]arcsinu+C
=2√x-2[√(1-x)]arcsin√x+C