Integrate it

$\text{Let }\displaystyle \int^{\pi}_{0} {\dfrac{x}{1+\sin{x}}\mathrm{d}x}=I$ $\text{By using the property }\boxed{\displaystyle\int _{ 0 }^{ 2a }{ f(x) } =\displaystyle\int _{ 0 }^{ a }{ f(x)\mathrm{d}x } +\displaystyle\int _{ 0 }^{ a }{ f(2a-x)\mathrm{d}x }} \text{ we get} \\ \Rightarrow I=\displaystyle\int^{2(\pi/2)}_{0} {\dfrac{x}{1+\sin{x}}\mathrm{d}x} =\displaystyle\int^{\pi/2}_{0}{\left(\dfrac{(\pi-x)}{1+\sin{(\pi-x)}}+\dfrac{x}{1+\sin{x}}\right)\mathrm{d}x}=\displaystyle \int^{\pi}_{0}{\dfrac{\pi}{1+\sin{x}}\mathrm{d}x}$ $\Rightarrow I= 2\pi \displaystyle \int^{\pi}_{0}{\frac{\sec^{2}{\left(\dfrac x2 \right)}}{\left(1+\tan{\left(\dfrac x2\right)}\right)^{2}}\mathrm{d}x}$ $\text{ Now, let } \left(1+\tan{\left(\dfrac x2 \right)}\right)=t \Rightarrow \mathrm{d}t=\sec^{2}{\left(\dfrac x2\right)}\mathrm{d}x$ $\Rightarrow I= 2\pi \displaystyle \int^{2}_{1}{\dfrac{1}{t^{2}}\mathrm{d}t }=2\pi\left[ -\dfrac{1}{t} \right]^{2}_{1}=2\pi \left(\dfrac12 \right)=\boxed{\pi}$

$\begin{aligned} I & = \int_0^\pi \frac x{1+\sin x} dx & \small \color{#3D99F6}{\text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b - x) \ dx} \\ & = \frac 12 \int_0^\pi \left(\frac x{1+\sin x} + \frac {\pi-x}{1+\color{#3D99F6}{\sin (\pi-x)}}\right) dx & \small \color{#3D99F6}{\text{Note that }\sin (\pi-x) = \sin x} \\ & = \frac 12 \int_0^\pi \frac \pi{1+\sin x} dx & \small \color{#3D99F6}{\text{Let } t = \tan \frac x2, \ \sin x = \frac {2t}{1+t^2}, \ dx = \frac {2 \ dt}{1+t^2}} \\ & = \frac \pi 2 \int_0^\infty \frac 2{(1+t^2)\left(1+\frac {2t}{1+t^2}\right)} dt \\ & = \pi \int_0^\infty \frac 1{(1+t)^2} dt \\ & = \frac \pi {1+t} \bigg|_\infty^0 = \pi \approx \boxed{3.1415} \end{aligned}$