ज्ञात करें `int ( x+ sin x)/( 1+ cos x) dx`

Explanation

`I=int (x+sin x)/(1+sin x) dx=int (x)/(1+cos x)dx+int(sin x)/(1+cos x)dx`
`=int x/(2cos^2""x/2)dx+int (2sin""x/2""cos ""x/2)/("2 cos"^2""x/2)dx=1/2intxsec^2 ""x/sdx+int tan""x/2dx`
`=1/2[x(ta n""x/2)/(1/2)-int 1*(tan""x/2)/(1/2)""dx]+int" "tan""x/2 -c`
`=x tan ""x/2-int" "tan""x/2dx +int " "tan ""x/2 = x tan ""x/2 +c`