∫x+sinx1+cosx dx is equal to :
(4)
Ans. x·tanx2+C
Explanation :
I=∫x+sinx1+cosx dx
=∫x1+cosx dx+∫sinx1+cosx dx
=∫x2cos2x2 dx+∫2sinx2cosx22cos2x2 dx
=12∫xsec2x2 dx+∫tanx2 dx
=12[x·tanx2·2-∫tanx2·2 dx]+∫tanx2 dx
=x·tanx2+C [By cancelling the last two integrals]