Question

Integrate $\int e^2\left(\frac{1+\sin x}{1+\cos x}\right)dx$

Answer 1

[A]$\int e^2\frac{(1+sinx)}{(1+cosx)}dx$

=$\int \frac{e^{2}sinx}{1+cosx}dx+\int \frac{e^2}{1+cosx}dx$

For$\int \frac{sinx}{1+cosx}dx$

Let $u=cosx+1\Rightarrow dx=\frac{-1}{sinx}du$

$=-\int \frac{1}{u}du=-ln\left(u\right)$

$For\int \frac{1}{1+cosx}dx$

$=\int \frac{se{c}^2\left(x12\right)}{2}dx$

Let $u=x12\Rightarrow dx=2du$

$=\int se{c}^{2}udu=tan=tan\frac{x}{2}$

so,

= $e^{2}tan\frac{x}{2}-e^{2}in(cosx+1)+c$

=$e^2[tan\frac{x}{2}-in(cosx+1\left)\right]+c$