`int(x+sinx)/(1+cosx) dx` is equal to

Options

A. `log|1+cosx|+C`
B. `log|x+sinx|+C`
C. `x-tanx//2 +C`
D. `x tan((x)/(2))+C`

Answer

D

Explanation

Let `l=int(x+sinx)/(1+cosx)dx`
`int(x)/(2cos^(2)x//2)dx+int(2sin x//2cos x//2)/(2cos^(2)x//2)dx`
`=(1)/(2)intx sec^(2).(x)/(2)dx+int tan.(x)/(2)dx`
On integrating by parts in first integral, we get
`=(1)/(2)[x int sec^(2).(x)/(2)dx-int[(d)/(dx)(x)intsec^(2).(x)/(2)dx]dx]+inttan.(x)/(2)dx`
`=(1)/(2)[x tan.(x)/(2).2-2 int tan.(x)/(2)dx]+int tan.(x)/(2)dx`
`=x tan .(x)/(2)-int tan.(x)/(2)dx+int tan.(x)/(2)dx`
`=x tan.(x)/(2)+C`