求不定积分sen(X)((1+(cos(X))^2)^1/2)

JIUFA4092022-10-04 11:39:541条回答

求它的不定积分
麻烦写下过程

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yuegang660 共回答了17个问题 | 采纳率100%: 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; 1年前