Question

$\int \frac{e^x\left(1+sinx\right)dx}{1+cosx}=e^{x}f\left(x\right)+C$, 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)

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

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

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

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

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

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