<s>[INST] Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me. 
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du  $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty  }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
 [/INST] $\big[$Another (standard) method$\big]$ 
When we have integrals of the form: $\displaystyle \int \frac{dx}{a+b\sin (x) + c\cos (x)} $ we can perform the substitution: $t=\tan (x/2)$. Then, we have: $\displaystyle \sin(x) = \frac{2t}{1+t^2} $ , $\displaystyle \cos (x) = \frac{1-t^2}{1+t^2} $ , $\displaystyle dx= \frac{2dt}{1+t^2}$. 
First we must remove the sine term from the numerator. 
$\displaystyle \int_0^{\pi} \frac{1-\sin(x) -1 + 1}{\sin(x) + 1} dx = \int_0^{\pi} \frac{2}{\sin(x)+1} -1 dx$ 
Now we substitute. Also @$x=0 \rightarrow t=0$ and @$x=\pi \rightarrow t=\infty$. Thus:
$\displaystyle \int_0^{\infty} \Bigg( \frac{2}{\frac{2t}{1+t^2} +1} -1 \Bigg) \cdot \frac{2}{1+t^2} dt = \int_0^{\infty} \frac{4}{2t+1+t^2} - \frac{2}{1+t^2} dt = $ 
$\displaystyle = \int_0^{\infty} \frac{4}{(t+1)^2} dt - 2\tan^{-1}(t)\big|_0^{\infty} = 4 \left[ -\frac{1}{t+1} \right]_0^{\infty} -2\frac{\pi}{2} = -4 \left( \frac{1}{\infty} - \frac{1}{0+1} \right) -\pi = 4-\pi $
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