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Integration by $u$ - substitution works alright for me, like in the following integral:

$$\int \frac{1}{3x+5}dx\ \ \ \ \ \ \ \ \ \ $$

Making $u$ = $3x+5$ is valid because I'm defining $u$ as the function, and once I change the integrand I see no problem understanding the thoughts behind this. $u$ has no implicit domain before defining it. However, trig substitution integration makes me a bit more confused.

enter image description here $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

My question can be demonstrated in the integral above. The correct substitution here is: $$x = \frac{\sin\ \theta}{5} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ The substitution itself, the act of making this statement, can cause me some confusion. I know that's what I have to do so I can deal with the integral, but I don't know what that imples. Due to the nature of this function, $x$ is restricted to $-1/5< x < 1/5$. Previously, I thought I had my question answered when I asked a question of this nature here. The answer was comprehensive and informative, but I thought that this logic should apply to all integration by trig substitution integrals. Since $\sqrt{1-x^2}$ and $\sin\ y$ both have the same domain, assuming $x$ and $y$ are real numbers this is totally valid to substitute. However, the domains of my substitution and the function are different here, as the domain of $x= \frac{\sin\ \theta}{5}$ is valid for all real numbers. Granted, the range of $(2)$ is almost the domain of $(1)$ except $(2)$ is inclusive $[1/5,1/5]$. So what gives? If they have different domains, why am I allowed to do this?

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    $\begingroup$ It's continuous map, and you could constrain the domain of $\theta$ to let $\sin\theta/5$ have the same range as $x$ $\endgroup$ – Yujie Zha Jun 28 '17 at 0:29
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When solving a definite integral by substitution, remember that the limits of integration change along with the changed integral.

So the integral

$$ \int_{-1/5}^{1/5}\frac{dx}{\sqrt{1-25x^2}}$$

with the substitution $x=\dfrac{\sin y}{5}$ the limits $x=-\frac{1}{5}$ and $x=\frac{1}{5}$ first become $\sin y=-1$ and $\sin y=1$, and in the integral the limits of integration become $y=\sin^{-1}(-1)=-\frac{\pi}{2}$ and $y=\sin^{-1}(1)=\frac{\pi}{2}$ so we get the transformed integral

$$ \frac{1}{5}\int_{-\pi/2}^{\pi/2}\frac{\cos y \,dy}{\sqrt{1-\sin^2y}}$$

So the bounds on $\sin y$ are as they should be, between $-1$ and $1$.

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  • $\begingroup$ I see, so the two integrals are exactly the same thing? Thus an equality is held and everything is fine? $\endgroup$ – sangstar Jun 28 '17 at 1:52
  • $\begingroup$ That's correct. $\endgroup$ – John Wayland Bales Jun 28 '17 at 2:04

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