Definite Integral Problem using substitution $t=\tan(x/2)$

Question

Solve $$ \int_0^\pi \frac {x} {1+\sin x} dx $$ using $\sin x = \frac {2\tan(x/2)} {1+\tan^2(x/2)}$ and the substitution $t = \tan(x/2)$. I tried doing this but I got to a point where my integral limits were $0$ to $\infty $. This happened when I substituted for $\tan(x/2)$ Is there a way of doing this using this substitution only? And also why does this happen?

UPDATE: What I did - $$ I = \int_0^\pi \frac {\pi-x} {1+\sin x} dx = \int_0^\pi \frac {\pi} {1+ \sin x} dx - I$$ Using $\sin x = \frac {2\tan(x/2)} {1+\tan^2(x/2)}$ $$ $$ $$ 2I = \int_0^\pi \frac {\pi} {1+\sin x} dx = \pi\int_0^\pi \frac {\sec^2(x/2)} {1+ \tan^2(x/2)+2 \tan(x/2)} dx$$ Now if there were no limits, this could've been solved easily by $t = \tan x$. But I can't do that because if I did, the limits would become $0$ to $\infty$. A way to solve this would be multiplying and dividing by $1+ \sin x$ but I don't want to do that. I want to use $t=\tan x$

Answer 1

After the main symmetry trick, another symmetry trick and a rationalization: $$ \int_{0}^{\pi}\frac{du}{1+\sin u}=2\int_{0}^{\pi/2}\frac{1-\sin u}{\cos^2 u}\,du =2\left[\tan u-\frac{1}{\cos u}\right]_{0}^{\pi/2}=\color{red}{\large2}.$$

Answer 2

If we must solve the problem with the suggested substitution, the integral becomes $\int_0^\infty\frac{4\arctan t dt}{(1+t)^2}$. The identity $\int_0^\infty f(t)dt=\int_0^1\left(f(t)+\tfrac{1}{t^2}f\left(\tfrac{1}{t}\right)\right)dt$ changes this to $\int_0^1\frac{2\pi dt}{(1+t)^2}=\pi$.

Answer 3

Hint. Alternatively, by the change of variable $u=\pi-x$, one gets $$ I=\int_0^\pi \frac {x} {1+\sin x} dx=\int_0^\pi \frac {\pi-u} {1+\sin (\pi-u)} du=\pi\int_0^\pi \frac {1} {1+\sin u} du-I $$ the latter integral being easier to evaluate.