Question

Evaluate: $\int \frac{\sin x}{1+\sin x}dx$

Answer 1

GE: $\int \frac{sinx}{1+sinx}dx$

$\int \frac{sinx+1-1}{sinx+1}dx$

$=\int \frac{sinx+1}{sinx+1}dx-\int \frac{dx}{1+sinx}$

$\Rightarrow \int dx-\int \frac{dx}{1+sinx}$

$\Rightarrow x-\int \frac{dx}{1+2sin\frac{x}{2}cos\frac{x}{2}}$

$1+2sin\frac{x}{2}cos\frac{x}{2}={sin}^2\frac{x}{2}+{cos}^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}$

$={(sin\frac{x}{2}+cos\frac{x}{2})}^2$

$=\frac{{(tan\frac{x}{2}+1)}^2}{{sec}^2\frac{x}{2}}$

$=x-\int \frac{{sec}^{2}x/2dx}{{(tan\frac{x}{2}+1)}^2}$

Putting $tan\frac{x}{2}+1=t$

$\left({sec}^2\frac{x}{2}\right)\times \frac{1}{2}dx=dt$

$=x-2\int \frac{dt}{t^2}=x+\frac{2}{t}+c$

$=x+\frac{2}{(tan\frac{x}{2}+1)}+c$