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 $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
 A: Substitute $x=\pi/2-2t$ so the integral becomes
$$
-2\int_{\pi/4}^{-\pi/4}\frac{1-\cos 2t}{1+\cos 2t}\,dt=
2\int_{-\pi/4}^{\pi/4}\frac{1-\cos^2t}{\cos^2t}\,dt
=2\Bigl[\tan t - t\Bigr]_{-\pi/4}^{\pi/4}
$$
A: Hint:  Note that by symmetry our integral is twice the integral from $0$ to $\pi/2$. Then by symmetry replace $\sin x$ by $\cos x$. Then use the identities $\cos x=2\cos^2 (x/2)-1=1-2\sin^2(x/2)$.
A: \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}
A: The question asked "is there another way", so here's another one. Start by removing the trig from the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}}{1+\sin{x}} \, \mathrm{d}x = 
\int_0^{\pi} \frac{2 - (1 +\sin{x})}{1+\sin{x}} \, \mathrm{d}x = 
\int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1  \right) \, \mathrm{d}x = 
-\pi + \int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} $$
There are many ways to solve this but I would like to demonstrate the substitution $u = 1 + \sin x$. It would be much more convenient if the limits ran from $0$ to $\frac{\pi}{2}$, so that $x = \sin^{-1} (u-1)$ is a continuous, monotonic function from $[1,2]$ to $[0,\frac{\pi}{2}]$, and also because $\cos x$ and $\sin x$ are both positive for $0 \leq x \leq \frac{\pi}{2}$. The latter property is often useful so we can write $\cos x=\sqrt{1-\sin^2 x}$ and $\sin x=\sqrt{1-\cos^2 x}$. Fortunately symmetry considerations allow a change of limits:
$$\int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} =
\int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}}$$
We have $\frac{\mathrm{d}u}{\mathrm{d}x} = \cos x = \sqrt{1-\sin^2 x} =  \sqrt{1-(u-1)^2} = \sqrt{2u - u^2}$ so the integral becomes:
$$\int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}} =
\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{2u - u^2}} =
\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}}$$
This might look fearsome but is actually very amenable to the third Euler substitution, since we have a factorised quadratic expression inside the square root. The computation is very similar to this answer. In general the substitution is $\sqrt{au^2 + bu + c} = \sqrt{a(u-\alpha)(u-\beta)} = (u-\alpha)t$ which gives $u = \frac{a\beta-\alpha t^2}{a-t^2}$; in our case we can take $a=-1$, $b=2$, $c=0$, $\alpha=0$, and $\beta=2$ with $\sqrt{-u(u-2)} = ut$.
Since $u=\frac{(-1)(2)-0t^2}{-1-t^2}=2(t^2+1)^{-1}$ we obtain $\frac{\mathrm{d}u}{\mathrm{d}t} = -4t(t^2+1)^{-2}$ and to change the limits we set $t=\frac{\sqrt{-u(u-2)}}{u} = \sqrt{\frac{2-u}{u}}$. The remaining integral becomes:
$$\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}} = 
\int_1^0 \frac{4 (-4t)(t^2+1)^{-2} \, \mathrm{d}t}{u^2 t} =
\int_0^1 \frac{4 (4t)(t^2+1)^{-2} \, \mathrm{d}t}{4(t^2+1)^{-2} t} =
\int_0^1 4 \, \mathrm{d}t =
4$$
This was not the most straightforward way to find the result $4 - \pi$, but I just wanted to draw out the similarity between this integral and one the original poster had asked about before (the main differences being the trig in the numerator - which is easily removed - and the limits).
A: Contour integration is quite a good idea in this case. On the unit circle, we have
$$z=e^{ix}=\cos x + i \sin x$$
and therefore
$$\sin x=\frac{z-z^{-1}}{2i}$$
and
$$dz=iz\,dx$$
Substitution makes the integral
$$I=\int_C \frac{1}{zi}\frac{-z^2+2iz+1}{z^2+2iz-1}dz=-\int_C\frac{1}{zi}\frac{(z-i)^2}{(z+i)^2}dz=$$
$$=i\int_C\frac{1}{z}\left(\frac{z^2-1-2zi}{z^2+1}\right)^2dz$$
where the integration contour $C$ is the upper half of the unit arc (clockwise). Now you can consider integration around the upper half of the unit disc. The integration splits symbolically as
$$\text{residue}=\int_{-1}^1+\int_C$$
The residue unfortunately lies directly on the integration path, at $z=0$, and is equal to $\operatorname{Res}=i$. As the path is straight at the residue, you can take the Cauchy principal value of the integral and use half the residue:
$$I=\frac12 2\pi i (i)-i\int_{-1}^1 \frac{1}{z}\left(\frac{z^2-1-2zi}{z^2+1}\right)^2dz$$
$$=-\pi-i\int_{-1}^1 \frac{1}{z}\left(\frac{(z^2-1)^2-4z^2-4zi(z^2-1)}{(z^2+1)^2}\right)dz$$
We keep only the even terms under the integral (this is the "Cauchy principal value" step):
$$I=-\pi-4\int_{-1}^1 \frac{(z^2-1)}{(z^2+1)^2}dz \quad (*)$$
$$I=-\pi-4\int_{-1}^1 \left(\frac{1}{z^2+1}-\frac{2}{(z^2+1)^2}\right)dz$$
These integrals are elementary (or tabulated in the case of the last term) and lead to the result
$$I=-\pi-4(\pi/2-2\cdot 1/2(1+\pi/2)))=4-\pi$$
Essentially, the only integral to compute is the one marked $(*)$, the rest is just simplification of rational functions. The integral seems to be more trivial than it looks (splitting it as I have is apparently not the most optimal solution, as half of it cancels out). If anyone notices a clever trick to get $\int (x^2-1)/(1+x^2)^2dx=-x/(1+x^2)$ (apart from exclaiming "it's obvious now" when you see the result), let me know in the comments.
A: Yet another way. Using $u=\tan(x)$, we get
$$
\begin{align}
\int\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x
&=\int\lower{2pt}{\frac{1-\frac{u}{\sqrt{1+u^2}}}{1+\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}}\\
&=\int\left(1-\raise{2pt}{\frac{u}{\sqrt{1+u^2}}}\right)^{\!\!2}\,\mathrm{d}u\\
&=\int\left(2-\raise{2pt}{\frac{2u}{\sqrt{1+u^2}}}-\frac1{1+u^2}\right)\,\mathrm{d}u\\[6pt]
&=2u-2\sqrt{1+u^2}-\arctan(u)+C\\[14pt]
&=2\tan(x)-2\sec(x)-x+C
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\pi\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x
&=(2-\pi)-(-2)\\[6pt]
&=4-\pi
\end{align}
$$
A: $\big[$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 $
A: First use partial fractions to get rid of the sine in the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$
We have the identity
$$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} $$
(from $\cos{2\theta}=1-2\sin^2{\theta}$ and $\sin{\theta}=-\cos{(\theta+\pi/2)}$), so the remaining integral is
$$ \int_0^{\pi} \csc^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \, dx = \left[ -2 \cot{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \right]_0^{\pi} = -2(-1-1) = 4 $$
A: Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $
$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$
We know that $\tan^2x + 1 = \sec^2x$. So the integral would look like: 
$$= \int_0^\pi 2\sec^2x -1 - 2\dfrac{\sin x}{\cos^2x}dx$$
We know the integral of $\sec^2x$ is just $\tan x$, and the last integrand can be solved by letting $u=\cos x$. 
