Question

$\int\frac{e^{x}\left(1+sin x\right)dx}{1+cos x} = e^{x }f \left(x\right) + C$, then f {x) is equal to

• A. sin $\frac{x}{2}$
• B. cos $\frac{x}{2}$
• C. tan $\frac{x}{2}$
• D. log $\frac{x}{2}$

Answer 1

Answer: (C)

$\int\frac{e^{x}\left(1+sin x\right)dx}{1+cos x} $

$= \int e^{x}\left[\frac{1}{2}sec^{2} \frac{x}{2} +tan \frac{x}{2}\right]dx$

$ =\frac{1}{2}\int e^{x} sec^{2} \frac{x}{2} dx +\int e^{x} tan \frac{x}{2} dx $

$ = e^{x} \frac{x}{2} -\int e^{x} tan \frac{x}{2} dx +\int e^{x} tan \frac{x}{2} dx +C = e^{x}tan \frac{x}{2} +C$

But $I =e^{x} f \left(x\right) + C$

$\therefore f \left(x\right) = tan \frac{x}{2}$