Question

Integrate the following functions.
$\int \frac{sinx}{1+cosx}dx.$

Answer 1

$\int \frac{sinx}{1+cosx}dx.$
Let $I=\int \frac{sinx}{1+cosx}dx$
Let $1+cosx=t\Rightarrow -sinx=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{-sinx}$
$\therefore I=\int \frac{sinx}{t}\times \frac{dt}{-sinx}=-\int \frac{1}{t}dt=-log|t|+C=-log|1+cosx|+C=log\left|\frac{1}{1+cosx}\right|+C(\because -loga=log\frac{1}{a})$