(tanx)^1/2的不定积分?
展开全部
t=(tanx)^(1/2),dx=2tdt/(1+t^4)
原式=St*2tdt/(1+t^4)=2S(1+t^2)/(1+t^4) dt-2S1/(1+t^4)dt
=根2*arctan(t-1/t)-2ln|t| + (1/2)ln(t^4+1) + C
t=(tanx)^(1/2)代入化简即可.,9,1/(2√2) (
-2 ArcTan[1 - √2 (Tan[x])^(1/2) ] + 2 ArcTan[1 + √2 (Tan[x])^(1/2) ]
+ Log[ -1 + √2 (Tan[x])^(1/2) - Tan[x] ] - Log[ 1 + √2 (Tan[x])^(1/2) + Tan[x] ]
) + C,2,
原式=St*2tdt/(1+t^4)=2S(1+t^2)/(1+t^4) dt-2S1/(1+t^4)dt
=根2*arctan(t-1/t)-2ln|t| + (1/2)ln(t^4+1) + C
t=(tanx)^(1/2)代入化简即可.,9,1/(2√2) (
-2 ArcTan[1 - √2 (Tan[x])^(1/2) ] + 2 ArcTan[1 + √2 (Tan[x])^(1/2) ]
+ Log[ -1 + √2 (Tan[x])^(1/2) - Tan[x] ] - Log[ 1 + √2 (Tan[x])^(1/2) + Tan[x] ]
) + C,2,
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询