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

Explanation

`int (e^x (1+sinx))/(1+cosx)dx=inte^x (1/(1+cos x)+(sinx)/(1+cosx))dx`
`=int e^x (1/(2cos^2""x/2+(2sin""x/2"cos"x/2)/(2cos^2""x/2)dx = int e^x (1/2 sec^2""x/2 + tan ""x/2)))dx`
`=int e^x [ f'(x)+f(x)]dx,` जहां `f(X)=tan""x/2`
`f^x f(x) +c c^x tan ""x/2 +c`