Question

Integrate the function: $e^x(\frac{1+sinx}{1+cosx})$

Answer 1

The correct answer is: $e^xtan\frac{x}{2}+C$
$e^x(\frac{1+sinx}{1+cosx})$
$=\bigg(\frac{e^x\frac{sin^2x}{2}+\frac{cos^2x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}\bigg)$
$=\frac{e^x(sin\frac{x}{2}+cos\frac{x}{2})^2}{2cos^2\frac{x}{2}}$
$=\frac{1}{2}e^x.\bigg(\frac{(sin\frac{x}{2}+cos\frac{x}{2})}{(cos\frac{x}{2})}\bigg)^2$
$=\frac{1}{2}e^x[tan\frac{x}{2}+1]^2$
$=\frac{1}{2}e^2(1+tan\frac{x}{2})^2$
$=\frac{1}{2}e^x[1+tan^2\frac{x}{2}+2tan\frac{x}{2}]$
$=\frac{1}{2}e^x[sec^2\frac{x}{2}+2tan\frac{x}{2}]$
$\frac{e^x(1+sinx)dx}{(1+cosx)}=e^x[\frac{1}{2}sec^2\frac{x}{2}+tan\frac{x}{2}]...(1)$
Let $tan\frac{x}{2}=ƒ(x) \,\,ƒ'(x)=\frac{1}{2}sec^2 \frac{x}{2}$
It is known that,$∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C$
From equation(1),we obtain
$∫\frac{e^x(1+sinx)}{(1+cosx)}dx=e^xtan\frac{x}{2}+C$