I like this integral too

Using tangent half-angle substitution , $t = \tan(\frac{x}{2})$ , we get $\begin{aligned} \int_0^{\pi}\frac{1}{1+\sin(x)}\,dx = & \int_0^{\infty} \frac{1}{1+\frac{2t}{1+t^2}}\frac{2}{1+t^2}\,dt \\ = & 2\int_0^{\infty}\frac{1}{1+2t+t^2}\,dt \\ = &2 \int_0^{\infty} \frac{1}{(1+t)^2}\,dt \\ = & -\frac{2}{1+t}\Big|_0^{\infty} \\ = & \lim_{t\to\infty}\frac{-2}{1+t} - \big(\frac{-2}{1}\big) \\ = & \boxed{2} \end{aligned}$

Note first that $\dfrac{1}{1 + \sin(x)} = \dfrac{1}{1 + \sin(x)} \times \dfrac{1 - \sin(x)}{1 - \sin(x)} = \dfrac{1 - \sin(x)}{1 - \sin^{2}(x)} = \dfrac{1 - \sin(x)}{\cos^{2}(x)} = \sec^{2}(x) - \dfrac{\sin(x)}{\cos^{2}(x)}$ .

Now the integral of $\sec^{2}(x)$ is $\tan(x)$ , and for the $\dfrac{\sin(x)}{\cos^{2}(x)}$ term we can use the substitution method with $u = \cos(x) \Longrightarrow du = -\sin(x) dx$ , giving us that

$\displaystyle \int_{0}^{\pi} \dfrac{1}{1 + \sin(x)} dx = \left(\tan(x) - \dfrac{1}{\cos(x)}\right)_{0}^{\pi} = (0 - (-1) - (0 - 1) = \boxed{2}$ .