高数.. 求不定积分 ∫1/cos2x√tgx
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令 √tanx= u，则 tanx = u^2, x = arctan(u^2), dx =2udu/(1+u^4),
cos2x = (cosx)^2 - (sinx)^2 = 1/(1+u^4) - u^4/(1+u^4) = (1-u^4)/(1+u^4)
I = ∫[1/(cos2x√tanx)]dx = ∫[2udu/(1+u^4)]/[u(1-u^4)/(1+u^4)]
= ∫2du/(1-u^4) = ∫[1/(1-u^2)+1/(1+u^2)]du
= ∫[(1/2)/(1-u)+(1/2)/(1+u)+1/(1+u^2)]du
= -(1/2)ln|1-u| + (1/2)ln|1+u| + arctanu + C
= arctanu + (1/2)ln|(1+u)/(1-u)| + C
= arctan√tanx + (1/2)ln|(1+√tanx)/(1-√tanx)| + C
cos2x = (cosx)^2 - (sinx)^2 = 1/(1+u^4) - u^4/(1+u^4) = (1-u^4)/(1+u^4)
I = ∫[1/(cos2x√tanx)]dx = ∫[2udu/(1+u^4)]/[u(1-u^4)/(1+u^4)]
= ∫2du/(1-u^4) = ∫[1/(1-u^2)+1/(1+u^2)]du
= ∫[(1/2)/(1-u)+(1/2)/(1+u)+1/(1+u^2)]du
= -(1/2)ln|1-u| + (1/2)ln|1+u| + arctanu + C
= arctanu + (1/2)ln|(1+u)/(1-u)| + C
= arctan√tanx + (1/2)ln|(1+√tanx)/(1-√tanx)| + C
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