##\int \frac{1}{\sqrt{1-x^2}} dx##

[Bashyboy]

Homework Statement

I am doing a little review and having a some trouble deriving the integral $\int \frac{1}{\sqrt{1-x^2}} dx$

Homework Equations

The Attempt at a Solution

Initially I was trying to solve this integral using the substitution $\cos \theta = x$. I drew my triangle so that the side adjacent to $\theta$ was $x$, the hypotenuse was $1$, and from this found that the opposite side was $\sqrt{1-x^2}$; after this I computed my differentials, performed various substitutions, and concluded that $\int \frac{1}{\sqrt{1-x^2}} dx = - \arccos x + c$. However, this wrong. So, I tried the substitution $\sin \theta = x$ and I derived the correct formula. Why didn't my first substitution work?

 

[robphy]

http://www.wolframalpha.com/input/?i=diff(-arccos(x),x)
http://www.wolframalpha.com/input/?i=diff(arcsin(x),x)

 

[Ray Vickson]

Homework Statement

I am doing a little review and having a some trouble deriving the integral $\int \frac{1}{\sqrt{1-x^2}} dx$

Homework Equations

The Attempt at a Solution

Initially I was trying to solve this integral using the substitution $\cos \theta = x$. I drew my triangle so that the side adjacent to $\theta$ was $x$, the hypotenuse was $1$, and from this found that the opposite side was $\sqrt{1-x^2}$; after this I computed my differentials, performed various substitutions, and concluded that $\int \frac{1}{\sqrt{1-x^2}} dx = - \arccos x + c$. However, this wrong. So, I tried the substitution $\sin \theta = x$ and I derived the correct formula. Why didn't my first substitution work?

Your first expression $-\arccos(x) + C$ is correct; differentiate it to check. That is something that you should always do.

Think about why there are two different-looking answers!

 

[vela]

If you can't figure out what Ray's getting at, try this: set C=0 for simplicity and plot the two results.

 

[robphy]

Call the integral $\theta\equiv \int \frac{1}{\sqrt{1-x^2}} dx$.
Then we have
$\theta=-\arccos(x) + C_1$ and $\theta=\arcsin(x) + C_2$,

Something that is sometimes forgotten is that the constants of integration ($C_1$ and $C_2$ ) are not generally equal.
That is, if we choose a value for $C_1$, what is the corresponding choice for $C_2$?
Let $C_2=C_1+Q$, where $Q$ is an unknown.

In this case, it might help interpreting the results as follows:
$\cos(C_1-\theta)=x$ and $\sin(\theta-C_2)=x$.
What is $Q$?

 

[robphy]

http://www.wolframalpha.com/input/?i=diff(-arccos(x),x)
http://www.wolframalpha.com/input/?i=diff(arcsin(x),x)

and here's another:
http://www.wolframalpha.com/input/?i=diff(2*arcsin(sqrt((x+1)/2)),x)
http://www.wolframalpha.com/input/?i=diff(-2*arccos(sqrt((x+1)/2)),x)