tanx乘以根号下1+sec^4的积分是什么?

Question

tanx乘以根号下1+sec^4的积分是什么?

Answer 1

I = ∫tanx√[1+(secx)^4]dx
= ∫{(secx)^2tanx√[1+(secx)^4]/(secx)^2}dx
= ∫{tanx√[1+(secx)^4]/(secx)^2}dtanx
= (1/2)∫{√[1+(secx)^4]/(secx)^2}d(tanx)^2
= (1/2)∫{√[1+(secx)^4]/(secx)^2}d[1+(tanx)^2]
= (1/2)∫{√[1+(secx)^4]/(secx)^2}d(secx)^2 [令 (secx)^2=u]
= (1/2)∫{√[1+u^2]/u}du [令 u=tanv ]
= (1/2)∫[secv/tanv]dtanv = (1/2)∫cscvdtanv
= (1/2)[cscvtanv +∫tanvcscvcotvdv]
= (1/2)[secv + ∫cscvdv] = (1/2)[secv + ln|cscv-cotv|] +C
= (1/2)√(1+u^2)+(1/2)ln|[√(1+u^2)-1]/u| +C
= (1/2)√[1+(secx)^4]+(1/2)ln|{√[1+(secx)^4]-1}/(secx)^2| +C