Why does this bound of integration become pi/4?

Question

Perhaps it is something easy that I am not seeing but I cannot seem to understand why, in this integral, a changes to $\frac\pi4$

3$\int_0^a$$\frac {dx}{\sqrt {a^2+x^2}}$

I was able to solve the integral but am stuck on why I needed to change a to $\frac\pi4$. Is it assumed a=1? Thanks

Edit: This integral is solved using trig substitution. My question was only about why the upper bound a is changed to $\frac\pi4$ instead of another value after doing the trig sub. The question was answered by Travis below.

Answer 1

Presumably the substitution made in the given solution is the standard trigonometric substitution $$x = a \tan \theta, \qquad dx = a \sec^2 \theta \,d\theta ,$$ which eliminates the radical from the integrand. At the lower limit, we have $x = 0$, and since $\theta = 0$ solves $0 = a \tan \theta$, we can take the lower limit to be $0$. At the upper limit, we have $x = \frac{\pi}{4}$, and to write the upper limit with respect to $\theta$, we must solve $a = a \tan \theta$, or $\tan \theta = 1$, which has solution $\frac{\pi}{4}$. Hence, with respect to $\theta$, we write the integral as $$\int_0^{\pi / 4} \cdots d\theta .$$

(This explanation brushes under the rug a technical point, namely that once we've picked the lower limit, we must pick a suitable upper limit, that is, we cannot pick any solution to $\tan \theta = 1$ for the upper limit. Since $\tan \theta$ is increasing and continuous on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, and that interval contains the chosen lower limit $0$, we should take the unique solution to $\tan \theta = 1$ in that interval.)

Answer 2

Perhaps because a change of variable was done:

$$x=a\tan u\implies dx=\frac a{\cos^2u}du\;,\;\;\begin{cases}x=0\implies u=0\\{}\\x=a\implies\tan u=1\implies u=\frac\pi4\end{cases}$$

so that

$$3\int_0^a\frac{dx}{\sqrt{a^2+x^2}}=3\int_0^{\pi/4}\frac{du}{\cos^2u}\cdot\frac{1}{\sqrt{1+\tan^2u}}=3\int_0^{\pi/4}\frac1{\cos u}du=$$

$$=\frac32\int_0^{\pi/4}\left(\frac{\cos u}{1-\sin u}+\frac{\cos u}{1+\cos u}\right)du=\frac32\,\log\left|\frac{1+\sin u}{1-\sin u}\right|_0^{\pi/4}=...$$