求解这两个式子的答案
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/31 01:41:47
求解这两个式子的答案
1.∫e^9xsinxdx=1/9∫sinxde^9x
=1/9(e^9xsinx-∫e^9xdsinx)
=1/9e^9xsinx+1/9∫e^9xcosxdx
=1/9e^9xsinx+1/81∫cosxde^9x
=1/9e^9xsinx+1/81e^9xcosx+1/81∫e^9xsinxdx
so ∫e^9xsinxdx=9/80e^9xsinx+1/80e^9xcosx+c
分部积分法
2.set =√sint=u,u^2=sint,so t=arcsinu^2,and dt=[2u/(√1-u^4)]du
so that:∫cos^5tdt/√sint=2∫(1-u^2)^2du=2/9u^9-4/5u^5+2u+c
换元积分法
=1/9(e^9xsinx-∫e^9xdsinx)
=1/9e^9xsinx+1/9∫e^9xcosxdx
=1/9e^9xsinx+1/81∫cosxde^9x
=1/9e^9xsinx+1/81e^9xcosx+1/81∫e^9xsinxdx
so ∫e^9xsinxdx=9/80e^9xsinx+1/80e^9xcosx+c
分部积分法
2.set =√sint=u,u^2=sint,so t=arcsinu^2,and dt=[2u/(√1-u^4)]du
so that:∫cos^5tdt/√sint=2∫(1-u^2)^2du=2/9u^9-4/5u^5+2u+c
换元积分法