calculus - To Find $int_0^{pi} frac{sin(x)}{1+sin(x)}$ why the substitution of $sin(x)=t$ Gives wrong answer?

To find out this basic integral
$$\int_0^{\pi} \frac{\sin(x)}{1+\sin(x)} \,\mathrm{d}x$$
I though of two methods :

Method 1:

I started by multiplying and dividing by $1-\sin(x)$ and then manipulating it one easily gets -
$$\int_0^{\pi} {(\sec(x)\tan(x) - (\tan(x))^2})\,\mathrm{d}x$$
Which is quite easy to calculate and gives value of $\pi-2$
I do not have any problem with this method , even though it took me some time to solve it.

Method 2 :
This was first thing I had thought of :

To let $\sin(x)=t$ and then when I tried to change the limits of integral I found that this substitution makes both upper and lower limits as $t=0$ which would give The value of above integral = 0 , according to the property
$\int_a^a f(x)\,\mathrm{d}x = 0$.

But the previous method gives answer of $\pi-2$ then what is wrong with the method 2 . Is that substitution incorrect ? But how and why ?

Answer

The sine function is not an injective function over the interval $(0,\pi)$. If you want to apply the substitution $\sin(x)\mapsto z$, you have to break the integration range in halves: this because a valid substitution is given by a diffeomorphism, not just a differentiable map.

In simple words, you are allowed to state that
$$\int_{a}^{b}f(x)\,dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(s))\,g'(s)\,ds$$
only if $g$ is an injective function over the involved integration range, and $\sin(x)$ is not injective over $(0,\pi)$. Otherwise you would get $\int_{0}^{\pi}\sin(x)\,dx=0$ and that is clearly wrong.

A possible way to go is: since the $\sin(x)$ function is symmetric with respect to the point $x=\pi/2$,

$$ \int_{0}^{\pi}\frac{\sin(x)\,dx}{1+\sin(x)}=2\int_{0}^{\pi/2}\frac{\sin(x)\,dx}{1+\sin(x)}=2\int_{0}^{1}\frac{t\,dt}{(1+t)\sqrt{1-t^2}}.$$

That is correct, even if not the most efficient way for computing such integral.
A more efficient way is to set $x=2\arctan\frac{t}{2}$ (aka Weierstrass substitution) to get
$$16\int_{0}^{+\infty}\frac{t\,dt}{(4+t^2)(2+t)^2}$$
that can be tackled through partial fraction decomposition.