Doubt about substitution in$\int_{-1}^{1}\sqrt{1+x^2}dx$ I've tried this substitution in this integral
$$\int_{-1}^{1}\sqrt{1+x^2}dx$$
Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$\frac{1}{2}\int_{1}^{1}\sqrt{\frac{1+t}{t}}dt=0$$
Which is obviously wrong. I know that this integral can be done with integration by parts or hyperbolic substitution, I want to know why this happens. My idea is that $x^2$ isn't always invertible, it is only on $[0, +\infty)$, and that causes this problem with the interval of integration. 
Am I right?
Thanks for your time.
 A: 
Error, When you set this  $x^2=t$, so $x=\sqrt{t}$ it is wrong because 
  $$-1\le x \le 1$$
  that is $x=\sqrt{t}$ is not only positive but can be $x= -\sqrt{t}$

Trick to reach your method  see that, $\sqrt{1+x^2}$ is an even function therefore, 
$$\int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx$$
Now you are allowed to:
Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$ \int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{\frac{1+t}{t}}dt$$

But Rather you could easilyy set $x =  \sinh u \implies dx = \cosh u du$

and $\sqrt{1+x^2} =\sqrt{1+\sinh u^2} = \cosh u$
$$\int_{-1}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{1+x^2}dx  \\=2\int_{0}^{\sinh^{-1}(1)} \cosh^2 u du= \int_{0}^{\sinh^{-1}(1)} 1+\cosh2 u du\\=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1))    $$
Using $$\cosh2u = \cosh^2 u +\sinh^2 u~~~and~~~~ \cosh^2 u -\sinh^2 u = 1.$$
on the other hand we have, 
$$\sinh 2u = 2c\cosh u \sinh u  = 2\sinh u\sqrt{1+\sinh^2 u}$$ 
Using this we have $$ \frac12 \sinh(2\sinh^{-1}(1))  =\sinh (\sinh^{-1}(1))\cdot\sqrt{1+\sinh^2\sinh^{-1}(1)} = \sqrt 2$$
Therefore, 
$$\color{red}{\int_{-1}^{1}\sqrt{1+x^2}dx  du=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1))  = \sinh^{-1}(1)+\sqrt2.} $$
A: The key question is this: As $x$ varies from $-1$ to $1$, what does $t$ do?
In many simple examples of $u$-substitution, the when $x$ varies across $(a,b)$ $u$ varies across $(c,d)$ and so we can just integrate the substituted function from $c$ to $d$ and be done with it.
We cannot do this in your case. In your case, the answer is that, as $x$ varies from $-1$ to $1$, $t$ varies from $1$ to $0$ and then back to $1$ again. Because there is this overlap, because $t$ traces the interval $[0,1]$ twice, you cannot simply replace the bounds. Additionally, on the two parts of this journey the function is actually different. For negative values of $x$ you need to use $-\sqrt{\quad}$ as the inverse function. This gives you:
$$\int_{-1}^1\sqrt{1+x^2}dx=\frac{1}{2}\int_1^0-\sqrt{\frac{1+t}{t}}dt+\frac{1}{2}\int_0^1\sqrt{\frac{1+t}{t}}dt=\int_0^1\sqrt{\frac{1+t}{t}}dt$$
Here the first integral is while $t$ is decreasing and the second is while $t$ is increasing. Notice that there's also a negative sign in front of this first integral. This is because $t$ is decreasing from $1$ to $0$ for values that correspond to negative $x$ values, so $x=-\sqrt{t}$ rather than $x=+\sqrt{t}$.
A: 
$$\int_{-1}^{1}\sqrt{1+x^2}\, \mathrm dx.$$
Let $x^2=t$, so $x=\sqrt{t},$ so we have
$\displaystyle\frac{1}{2}\int_{1}^{1}\sqrt{\frac{1+t}{t}}\,\mathrm dt=0,$ which is obviously wrong.

The given integration domain is $[-1,1];$ on $[-1,0],$ $$x^2=t\implies x=\color{red}{-\sqrt{t}}.$$ So, we need to split the integration interval into two: \begin{align}\int_{-1}^{1}\sqrt{1+x^2}\, \mathrm dx&=\int_{-1}^{0}\sqrt{1+x^2}\, \mathrm dx+\int_{0}^{1}\sqrt{1+x^2}\, \mathrm dx\\&=\int_{1}^{0}\frac{\sqrt{1+t}}{-2\sqrt t}\, \mathrm dt+\int_{0}^{1}\frac{\sqrt{1+t}}{2\sqrt t}\, \mathrm dt\\&=\int_{0}^{1}\sqrt{\frac{1+t}t}\, \mathrm dt,\end{align} which equals 2.2956.

My idea is that $x^2$ isn't always invertible, and that causes this problem with the interval of integration.

Integration by substitution does not inherently require injectivity. Point 4 of this linked answer contains a valid substitution $u=1+x^2,$ which, like the substitution $t=x^2$ here, is non-invertible.
