Question

高数大神求解20题

Answer 1

∫sin²x/cos³x dx
= ∫(sin²x/cos²x)(1/cosx) dx
= ∫tan²x*secx dx
= ∫(sec²x-1)*secx dx
= ∫sec³x dx - ∫secx dx
= I - J
J = ∫secx dx
= ∫secx*(secx+tanx)/(secx+tanx) dx
= ∫(secx*tanx+sec²x)/(secx+tanx)
= ∫d(secx+tanx)/(secx+tanx)
= ln|secx+tanx| + C
I = ∫sec³x dx
= ∫sec²x*secx dx
= ∫secx dtanx
= secx*tanx - ∫tanx d(secx)
= secx*tanx - ∫tan²x*secx dx
= secx*tanx - ∫(sec²x-1)*secx dx
= secx*tanx - ∫sec³x dx + ∫secx dx
= secx*tanx - I + J
2I = secx*tanx + J
I = (1/2)secx*tanx + (1/2)J
= (1/2)secx*tanx + (1/2)ln|secx+tanx|

原式= I - J
= (1/2)secx*tanx + (1/2)J - J
= (1/2)secx*tanx - (1/2)ln|secx+tanx| + C

追问

好麻烦

追答

没得法，有这么难