Integrate the expression $\int\frac{x}{1+\sin x}\ \text dx$

Question

Integrate the expression $\int\frac{x}{1+\sin x}\ \text dx$

I really do not see any identity working here. I tried rewriting in terms of $\cos x$ and got $$\int\frac{\frac{x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}\ \text dx$$

Answer 1

$$ \begin{align} \int\frac1{1+\sin(x)}\,\mathrm{d}x &=\int\frac{1-\sin(x)}{\cos^2(x)}\,\mathrm{d}x\\[6pt] &=\tan(x)-\sec(x)+C \end{align} $$ Therefore, we can integrate by parts $$ \begin{align} \int\frac{x}{1+\sin(x)}\,\mathrm{d}x &=\int x\,\mathrm{d}(\tan(x)-\sec(x))\\ &=x(\tan(x)-\sec(x))-\int(\tan(x)-\sec(x))\,\mathrm{d}x\\[6pt] &=x(\tan(x)-\sec(x))+\log(\cos(x))+\log(\sec(x)+\tan(x))+C\\[12pt] &=x(\tan(x)-\sec(x))+\log(1+\sin(x))+C\\[6pt] &=-\frac{x\cos(x)}{1+\sin(x)}+\log(1+\sin(x))+C \end{align} $$

Answer 2

$$I = \int\frac{x}{1+\sin x}\, dx$$ Multiply by the conjugate of $1 + \sin x$ $$\int\frac{x}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}\, dx = $$ $$\int\frac{x}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}\, dx = \int\frac{x-x\sin x}{1-\sin^2 x}\, dx$$ $$\int\frac{x-x\sin x}{\cos^2 x}\, dx$$ $$\int\frac{x}{\cos^2 x}\, dx- \int\frac{x\sin x}{\cos^2 x}\, dx$$ Apply Integration by Parts $$\int f(x)g(x)'\,dx = f(x)g(x) - \int f'(x)g(x)\, dx$$ where

$$\int\frac{x}{\cos^2 x}\, dx$$ $f(x) = x,\implies f'(x) = 1$
$g(x) = \tan x \implies g'(x) = \frac{1}{\cos^2x}$

$$x\tan(x) - \int\tan(x)$$ $$x\tan(x) + \ln(\cos x) +C\tag1$$
And $$\int\frac{x\sin x}{\cos^2 x}\, dx$$ $f(x) = x\sin x,\implies f'(x) = \sin x + x\cos x$
$g(x) = \tan x \implies g'(x) = \frac{1}{\cos^2x}\tan x$

$$x\sin x \tan x - \int(\sin x + x\cos x)\tan x\, dx$$ $$x\sin x \tan x - \int \sin x\tan x + x\cos x\tan x\, dx$$ $$x\sin x \tan x - \int x\sin x + \sin x\tan x\, dx$$ $$x\sin x \tan x - \left(\int x\sin x \, dx + \int \sin x\tan x\, dx\right)$$ $$x\sin x \tan x - \left(-x\cos x + \sin x + \int \sin x\tan x\, dx\right)$$
$\int\sin x\tan x\, dx = \int \frac{\sin^2 x}{\cos x}\, dx = \int \frac{1 - \cos^2 x}{\cos x}\, dx = \int \frac{1}{\cos x} - \cos x \, dx = \int \frac{1}{\cos x}\, dx - \int \cos x \, dx =$
$ \ln\left(\tan(x) + \sec(x)\right) - \sin(x) +C$


$$x\sin x \tan x - \left( -x\cos x + \sin x + \ln\left(\tan(x) + \sec(x)\right) - \sin(x)\right) + C$$ $$x\sin x \tan x + x\cos x - \ln\left(\tan(x) + \sec(x)\right) + C\tag2$$
Also...

$$I = \big(x\tan(x) + \ln(\cos x)\big) + \big(x\sin x \tan x + x\cos x - \ln\left(\tan(x) + \sec(x)\right)\big) + C$$

Answer 3

\begin{align} I&=\int\frac{xdx}{1+\sin x}=\int x d\left(\frac{-\cos x}{1+\sin x}\right)=\frac{-x\cos x}{1+\sin x}+\int \frac{\cos x}{1+\sin x}dx\\ &=\frac{-x\cos x}{1+\sin x}+\ln |1+\sin x|+C \end{align}