Validity of trigonometric substitutions

Question

Let us look at a term $\dfrac{x}{\sqrt{1+x^2}}$. Here $x>0$.

Now we can make a trigonometric substitution $x=\tan A$. But why does this $A$ have to be in $(0,\frac{\pi}{2})$? I don't understand this.

I am saying this from this video. https://youtu.be/VqoZLW05TOE

After 3.00 minutes,they say that all $A, B, C$ are within $(0,\frac{\pi}{2})$ which didn't make sense to me,the logic they gave beforehand is we have an isolated graph and they randomly chose $x$ on that graph,but $x$ can be outside of that range as well, so that seemed like a flawed explanation to me.

Answer 1

A substitution is often made to simplify an expression. The substitution $x=\tan A$ makes sense because both $x$ and $\tan A$ can assume values over the entire real line.

In the problem mentioned, $x$ is restricted to $R^+$. To restrict $\tan A$ to $R^+$ as well, we can restrict the domain of $\tan A$ to $(k\pi,(2k+1)\frac{\pi}{2}), k\in I$

The simplest of the domains that we could pick is $(0,\frac{\pi}{2})$, which of course keeps further calculations simple.

Answer 2

When making the above implicit substitution $x=\tan u,$ we are actually applying the change-of-variables theorem $\Bigg[$if $g'$ is integrable on $[a,b]$ and $f$ is continuous on $g[a,b],$ then $\displaystyle\int_a^bf\big(g(x)\big)g'(x)\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\,\mathrm{d}u\Bigg]$ with $u=g(x)$ and $x=\tan u,$ i.e., $$g=\arctan,$$ where $\arctan$ is notably an invertible function. The conventional choice of principal range for $\arctan$ is $\left(-\frac\pi2,\frac\pi2\right),$ although of course $\left(\frac\pi2,\frac{3\pi}2\right),$ for example, works too.

In your example, since $x=\tan u>0,$ we just need $u\in\left(0,\frac\pi2\right)\subset\left(-\frac\pi2,\frac\pi2\right)$.

I wrote more here: Does $u$-substitution require invertibility?