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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.

ACB
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green_blue
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  • if x is from -infinity to infinity, tanx will be from -pi/2 to pi/2. is you sub tan here, this in the root becomes 1/cos^2 t what makes the entire expression |cos t| *sint/cost if x < 0 you should change sign – user184868 Jan 26 '22 at 13:34
  • They have to, otherwise you get into problems with the substitution since then $$\sqrt{1+\tan^2x}=\frac1{\sqrt{\cos^2x}}=\frac1{|\cos x|}$$ – DonAntonio Jan 26 '22 at 13:35
  • But what if $x$ is a very large number for which there doesn't exist any $A$ within $(0,\frac{\pi}{2})$ – green_blue Jan 26 '22 at 13:46
  • @green_blue It is easy to see that $$0\le \frac x{\sqrt{1+x^2}}\le 1\ldots$$ – DonAntonio Jan 26 '22 at 13:58

2 Answers2

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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.

DatBoi
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  • But $x$ is a variable and can take large values as well. So what if there doesn't exist such an $A$ within $(0,\frac{\pi}{2})$ – green_blue Jan 26 '22 at 13:58
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    $\tan x$ spans the entire positive real line in $(0.\pi/2)$. Take a look at its graph. – DatBoi Jan 26 '22 at 14:00
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    @green_blue in fact they draw the graph for you starting at about 2:00. At 2:09 they say the word "asymptotes" referring to the vertical lines at $A=-\frac\pi2$ and $A=\frac\pi2$, which for the part of the graph near $A=\frac\pi2$ implies that it just keeps on going up and up and getting closer to $A=\frac\pi2$. Of course they only draw a finite amount of the curve because the whiteboard has finite height, but they assume you know the $\tan$ function and its properties. – David K Jan 26 '22 at 14:22
  • @DavidK yes! I would also suggest OP to revisit trigonometric functions! – DatBoi Jan 26 '22 at 14:25
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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?

ryang
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