Calculus 8 : Rational t'gos

$\begin{aligned} I & = \int_0^\pi \frac 1{1+\sin x} dx & \small \color{#3D99F6}{\text{Since }\sin x \text{ is symmetrical on }\frac \pi 2} \\ & = \color{#3D99F6}{2} \int_0^{\color{#3D99F6}{\frac \pi 2}} \frac 1{1+\sin x} dx & \small \color{#3D99F6}{\text{Let }t = \tan \frac x2, \ dt = \frac {1+t^2}2 dx, \ \sin x = \frac {2t}{1+t^2}} \\ & = 2 \int_0^\frac \pi 2 \frac 1{1+\frac {2t}{1+t^2}} dx \\ & = 2 \int_0^\frac \pi 2 \frac {1+t^2}{1+t^2 + 2t} dx \\ & = 4 \int_0^\frac \pi 4 \frac 1{(t+1)^2} dt \\ & = \frac 4{t+1} \ \bigg|^0_\frac \pi 4 \\ & = 4\left(1-\frac 12 \right) = \boxed{2} \end{aligned}$

Note that $\dfrac{1}{1 + \sin(x)} * \dfrac{1 - \sin(x)}{1 - \sin(x)} = \dfrac{1 - \sin(x)}{\cos^{2}(x)} = \sec^{2}(x) - \sec(x)\tan(x)$ ,

the integral of which is $\tan(x) - \sec(x)$ , which when evaluated from $x = 0$ to $x = \pi$ yields a value of

$(\tan(\pi) - \sec(\pi)) - (\tan(0) - \sec(0)) = (0 - (-1)) - (0 - 1) = 2 = 2^{1} \Longrightarrow \dfrac{a}{b} = \dfrac{2}{1} = \boxed{2}$ .