`int(1+sinx)/(sinx(1+cosx))dx`

Explanation

Let `I=int(1+sinx)/(sinx(1+cosx))dx`
Putting `sinx=(2tanx//2)/(1+tan^(2)x//2)` and, `cosx=(1-tan^(2)x//2)/(1+tan^(2)x//2)`
We get,
`I=int(1+(2tanx//2)/(1+tan^(2)x//2))/((2tanx//2)/(1+tan^(2)x//2)(1+(1-tan^(2)x//2)/(1+tan^(2)x//2)))dx = int((1+tan^(2)x//2+2tanx//2)(1+tan^(2)x//2))/(2tanx//2(1+tan^(2)x//2+1-tan^(2)x//2))`dx
`=int((1+tanx//2)^(2)sec^(2)x//2)/(4tanx//2)dx= int(1+t^(2)+2t)/(2t)`dt, where t`=tanx/2`
`=1/2int(1/t+t+2)dt=1/2[ln|t|+t^(2)/2+2t]+C=1/2[ln |tanx//2|(tan^(2)x//2)/(2) + 2tan(x//2)]+C`