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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.

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    $\begingroup$ $x= \pm \sqrt t$ $\endgroup$ – John Lou Oct 20 '17 at 18:51
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    $\begingroup$ Trig sub. Enjoy. $\endgroup$ – Randall Oct 20 '17 at 18:51
  • $\begingroup$ i would use $$x=\tan(t)$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 20 '17 at 18:52
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    $\begingroup$ First see that you function is even on a symmetric domain so break your integral in two $\endgroup$ – Guy Fsone Oct 20 '17 at 18:53
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    $\begingroup$ I think the OP is not asking how to do it, but why his method fails. $\endgroup$ – velut luna Oct 20 '17 at 18:56
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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, $$\color{red}{\int_{-1}^{1}\sqrt{1+x^2}dx du=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1)) = \sinh^{-1}(1)+\sqrt2.} $$

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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 $x=+\sqrt{t}$.

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