<s>[INST] Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework: 
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
 [/INST] If we choose to take a (purely) trigonometric route, we might start like this:
$$
\eqalign{
x&=\tan\theta\\
dx&=\sec^2\theta\,d\theta\\
I&=\int_0^1\sqrt{1+x^2}\,dx\\
&=\int_0^\frac{\pi}{4}\sec^3\theta\,d\theta\\
}
$$
However at this stage, we actually need the substitution
$t=\sec\theta+\tan\theta$, believe it or not:
$$
\eqalign{
t&=\sec\theta+\tan\theta\\
dt&=\left(\sec\theta\tan\theta+\sec^2\theta\right)\,d\theta
=t\sec\theta\,d\theta\\
\frac{dt}{t}&=\sec\theta\,d\theta
}
$$
And then we need some trigonometry inspirations:
$$
\eqalign{
t &
= \sec\theta+\tan\theta
= \frac{1+\sin\theta}{\cos\theta}
= \frac{\cos\theta}{1-\sin\theta}
\\
t-\frac1t &
= \frac{\cos\theta}{1-\sin\theta}
- \frac{\cos\theta}{1+\sin\theta}
= 2\tan\theta
\\
\tan\theta &
= \frac12 \left( t-\frac1t \right)
= \frac12 \left( t-t^{-1} \right)
\\
\tan^2\theta &
= \frac{t^2+t^{-2}}{4} - \frac12
\\
\sec^2\theta &
= \frac{t^2+t^{-2}}{4} + \frac12
}
$$
We can can then proceed as follows:
$$
\eqalign{
I &
= \int_0^\frac{\pi}{4}\sec^3\theta\,d\theta
= \int_1^{1+\sqrt{2}} \left( \frac{t^2+t^{-2}}{4}+\frac12 \right)\,\frac{dt}{t}
\\&
= \frac14 \int_1^{1+\sqrt{2}} \left(t+t^{-3}\right)\,dt
+ \frac12 \int_1^{1+\sqrt{2}} t^{-1}\,dt
\\&
= \frac18\left[t^2-t^{-2}\right]_1^{1+\sqrt2}
+ \frac12\left[\ln t \right]_1^{1+\sqrt2}
\\&
= \frac{\sqrt2+\ln{\left(1+\sqrt2\right)}}{2}
}
$$
where it is helpful to notice that
$$
\left(1+\sqrt2\right)^{-1} =
\frac{1}{1+\sqrt2} \cdot
\frac{1-\sqrt2}{1-\sqrt2} =
\frac{1-\sqrt2}{-1}
$$
so that
$$
\left(1+\sqrt2\right)^{-2} =
\left(1-\sqrt2\right)^2.
$$
</s>