- ∫x^2/(xsinx+cosx)^2dx积分
- 积分
- ∫x^2/(xsinx+cx)^2dx
因为:[1/(xsinx+cosx)]'=[0-(xsinx+cosx)']/(xsinx+cosx)^2
=[-(sinx+xcosx-sinx)]/(xsinx+cosx)^2
=-xcosx/(xsinx+cosx)^2
所以:
∫x^2/(xsinx+cosx)^2dx
=∫x^2/(-xcosx)d[1/(xsinx+cosx)]
=∫(-xsecx)d[1/(xsinx+cosx)]
=(-xsecx)/(xsinx+cosx)+∫[1/(xsinx+cosx)]d(xsecx)
=(-xsecx)/(xsinx+cosx)+∫[(secx+xsecxtanx)/(xsinx+cosx)]dx
=(-xsecx)/(xsinx+cosx)+∫[secx*(xsinx+cosx)/cosx]/(xsinx+cosx)dx
=(-xsecx)/(xsinx+cosx)+∫sec^2 xdx
=(-xsecx)/(xsinx+cosx)+tanx+C