Question

∫sinx/cosx^5dx，求不定积分

Answer 1

=∫-1/(cosx)^5d(cosx)
令cosx=t.则原式=
∫-1/t^5dt
=1/(4t^4)+c
反带回x
原式=
1/(4(cosx)^4)+c

Answer 2

=∫tanx(secx)^4dx
=∫sec³xdsecx
=(1/4)(secx)^4+C

追答

或=-∫1/(cosx)^5dcosx
=1/4(cosx)^4+C