Question

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

Answer 1

$\int e^x(\frac{1}{1+cosx}+\frac{sinx}{1+cosx})dx=\int e^x(\frac{1}{2co{s}^2\frac{x}{2}}+\frac{2sin\frac{x}{2}cos\frac{x}{2}}{2co{s}^2\frac{x}{2}})dx[\because cos2x=2co{s}^{2}x-1\text{and}sin2x=2sinxcosx]=\int e^x(\frac{se{c}^2\frac{x}{2}}{2}+tan\frac{x}{2})dx=\int e^x(tan\frac{x}{2}+\frac{1}{2}se{c}^2\frac{x}{2})dx$
Let $f\left(x\right)=tan\frac{x}{2}\Rightarrow f^{\prime }\left(x\right)=\frac{se{c}^2\frac{x}{2}}{2}$
$\therefore \int e^x(tan\frac{x}{2}+\frac{se{c}^2\frac{x}{2}}{2})dx=e^{x}tan\frac{x}{2}+C(\because \int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx=e^{x}f(x\left)\right)$