<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] \begin{align}
A &= \int_{0}^{\pi}\left(\:\frac{1\:-\:\sin (x)}{\sin (x)\:+\:1}\:\right)dx \\
F(x) &= \int\left(\:\frac{1\:-\:\sin (x)}{\sin(x)\:+\:1 }\cdot\frac{\sin(x)\:-\:1}{\sin (x)\:-\:1 }\:\right)dx \\
 &= \int\left(\:\frac{ -\sin^2(x) + 2\cdot sin(x)-1}{\sin^2(x)\:-\:1 }\:\right)dx \\
 &= \int\left(\:\frac{ -(\sin^2(x) - 2\cdot sin(x)+1)}{-(1-\sin^2(x)) }\:\right)dx \\
 &= \int\left(\:\frac{ (\sin^2(x) - 2\cdot sin(x)+1)}{(1-\sin^2(x)) }\:\right)dx \\
 &= \int\left(\:\frac{\sin^2(x) - 2\cdot sin(x)+1}{\cos^2(x) }\:\right)dx \\
 &= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}-\frac{  2\cdot sin(x)}{\cos^2(x)}+\frac{1}{\cos^2(x)}\:\right)dx \\
 &= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}\right)dx-\int\left(\frac{2\cdot sin(x)}{\cos^2(x)}\right)dx+\int\left(\frac{1}{\cos^2(x)}\:\right)dx \\
F(x) &= \tan(x) - x - 2\cdot\sec(x) + \tan(x)\\
A &= F(\pi) - F(0)\\
F(\pi) - F(0) &= (\tan(\pi) - \pi - 2\cdot\sec(\pi) + \tan(\pi))-(\tan(0) - 0 - 2\cdot\sec(0) + \tan(0))\\
  &= (0 - \pi - 2\cdot-1 + 0)-(0 - 0 - 2\cdot1 + 0)\\
  &= (-\pi + 2)-(-2)\\
  &= -\pi + 2+2\\
  &= 4-\pi\\
\end{align}
</s>