求不定积分sen(X)((1+(cos(X))^2)^1/2)
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/08/27 10:38:43
求不定积分sen(X)((1+(cos(X))^2)^1/2)
求它的不定积分
麻烦写下过程
![](http://img.wesiedu.com/upload/0/1c/01cab4b0285e2c4ec6b307152fb21726.jpg)
麻烦写下过程
I=-∫[1+cos(x)^2]^(1/2)dcos(x)
分部积分得I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫cos(x)d[1+cos(x)^2]^(1/2)
I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫{[cos(x)^2]/[1+cos(x)^2]^(1/2)}dcosx
I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫{[cos(x)^2+1-1]/[1+cos(x)^2]^(1/2)}dcosx
I=-cos(x)[1+cos(x)^(2)]^(1/2)-∫1/[1+cos(x)^2]^(1/2)}dcosx-I
2I=-cos(x)[1+cos(x)^(2)]^(1/2)-ln[(1+cos(x)^2]^(1/2)+cos(x)]
I={-cos(x)[1+cos(x)^(2)]^(1/2)-ln[(1+cos(x)^2]^(1/2)+cos(x)]}/2
分部积分得I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫cos(x)d[1+cos(x)^2]^(1/2)
I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫{[cos(x)^2]/[1+cos(x)^2]^(1/2)}dcosx
I=-cos(x)[1+cos(x)^(2)]^(1/2)+∫{[cos(x)^2+1-1]/[1+cos(x)^2]^(1/2)}dcosx
I=-cos(x)[1+cos(x)^(2)]^(1/2)-∫1/[1+cos(x)^2]^(1/2)}dcosx-I
2I=-cos(x)[1+cos(x)^(2)]^(1/2)-ln[(1+cos(x)^2]^(1/2)+cos(x)]
I={-cos(x)[1+cos(x)^(2)]^(1/2)-ln[(1+cos(x)^2]^(1/2)+cos(x)]}/2