3
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ It's continuous map, and you could constrain the domain of $\theta$ to let $\sin\theta/5$ have the same range as $x$ $\endgroup$ Commented Jun 28, 2017 at 0:29

1 Answer 1

0
$\begingroup$

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 $1$.

$\endgroup$
2
  • $\begingroup$ I see, so the two integrals are exactly the same thing? Thus an equality is held and everything is fine? $\endgroup$ Commented Jun 28, 2017 at 1:52
  • $\begingroup$ That's correct. $\endgroup$ Commented Jun 28, 2017 at 2:04

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.