# Integration by u substitution for inverse trig formulas

## Homework Statement

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?

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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

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

$\displaystyle \int\frac{\not{a}\frac{1}{a}}{a(1+\frac{x^2}{a^2})}dx$
I have crossed out an extra letter a , which you had in the numerator.
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?

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:

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

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

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:

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

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

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

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

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

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

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

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

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
...

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?
Yes, they are equivalent.

Thanks SammyS,

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:

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

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

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:
SammyS
Staff Emeritus
Homework Helper
Gold Member
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 $\frac{1}{a}\frac{1}{a}\frac{1}{(\frac{x}{a})^{2}+1}$ = $\frac{1}{x^{2}+a^{2}}$?
The algebra says yes.