Integration by u substitution for inverse trig formulas

1. Sep 28, 2012

LearninDaMath

1. The problem statement, all variables and given/known data

You know the U substitution proofs for inverse trig functions that go like this:

$\int\frac{1}{a^{2}+x^{2}}dx$

$\int\frac{a\frac{1}{a}}{a(1+\frac{x^2}{a^2})}dx$

let u = x/a

du= dx/a

....

$\frac{1}{a}tan^{-1}(x/a)+c$

I have searched google and can't find any of these proofs for the other trigonometric functions like sin and tan. Where can I find step by steps of these particular integration generalizations?

Do these U substitution methods for inverse trig functions have some sort of specific name?

2. Sep 28, 2012

HallsofIvy

Staff Emeritus
Last edited by a moderator: May 6, 2017
3. Sep 29, 2012

LearninDaMath

Hi Hallsofivy, really appreciate it. However, of the links you kindly provided, there is only one instance of a formula of which i'm describing. One for arcsin. But there seems to be no resource on the web that displays all or even 2 or 3 of these particular formulas. I've searched every combination of "inverse," "trigonometric," "integration," "subsititution," "formula(s)." I don't know how else to describe these specific formulas. Do you have any other keyword suggestions or links that provide on this integration topic?

4. Sep 29, 2012

SammyS

Staff Emeritus
I have crossed out an extra letter a , which you had in the numerator.
I don't see how what you have is much of a proof of anything.

You can get this from taking the derivative of tan-1(x/a).
$\displaystyle \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{1}{a}\frac{1}{(x/a)^2+1}$​

To do something similar with arcsin & arccos, go ahead and take the derivatives:

$\displaystyle \frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right)$

$\displaystyle \frac{d}{dx}\cos^{-1}\left(\frac{x}{a}\right)$

5. Sep 29, 2012

LearninDaMath

I didn't mean proof, sorry about that. I guess I just meant formula. Thanks, I'll give it a try.

6. Sep 29, 2012

LearninDaMath

$sin^{-1}\frac{x}{a}=y$

$\frac{d}{dx}sin^{-1}=\frac{d}{dx}\frac{x}{a}$

$cos(y)y'= \frac{1}{a}$

$y'=\frac{1}{acos(y)}$

$y'=\frac{1}{acos(sin^{-1}(\frac{x}{a})^{2})}$

Trig Identity: $sin^{2}y+cos^{2}y=1$

cosy=$\sqrt{1-(\frac{x}{a})^2}$

$y'=\frac{1}{acos(y)}=\frac{1}{a\sqrt{1-(\frac{x}{a})^{2}}}$

So when I do the derivative of arcsine, I get $\frac{d}{dx}sin^{-1}\frac{x}{a}=\frac{1}{a\sqrt{1-(\frac{x}{a})^{2}}}$

However, the formula states:

$\int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}\frac{x}{a}+C$

I seem to have gotten an "a" outside of the radical and a "1" inside the radical, while the formula shows nothing outside the radical and an "$a^{2}$ inside the radical. Where might I have gone off track to arrive at a final answer with misarranged variables?

EDIT:

I think I see it, would it just be a matter of distributing the "a" into the radical?

7. Sep 29, 2012

SammyS

Staff Emeritus
Yes, they are equivalent.

8. Sep 29, 2012

LearninDaMath

Thanks SammyS,

Also,

You show that the derivative of arctanx is:
$\displaystyle \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{1}{a}\frac{1}{(x/a)^2+1}$​

But if I am interested in deriving the function $\frac{1}{a}tan^{-1}(\frac{x}{a})$ for this particular formula, does the extra fraction $\frac{1}{a}$ that the function to be derived includes account for the final result of $\frac{1}{a}\frac{1}{a}\frac{1}{(\frac{x}{a})^{2}+1}$ = $\frac{1}{x^{2}+a^{2}}$?

Last edited: Sep 29, 2012
9. Sep 29, 2012

SammyS

Staff Emeritus
The algebra says yes.