# Trigonometric functions and integrals

1. Dec 7, 2016

### Any Help

1. The problem statement, all variables and given/known data
I'm searching for the integral that gives arcosu

2. Relevant equations
as we know : ∫u'/[1-u^2]^0.5 dx = arcsinu
derivative of arccosu = -u'/[1-u^2]^0.5 + C
derivative of arcsinu= u'/[1-u^2]^0.5
3. The attempt at a solution
when I type the -u'/[1-u^2]^0.5 on the online integral calculator it always gives me -arcsinu +C
Why it doesn't gives arccosu
is the arccosu=-arcsinu?!
and what is the integral that gives arcosu?

2. Dec 7, 2016

### cnh1995

cos-1u= -sin-1u+90°.
Sine and cosine functions have a phase difference of 90°.

3. Dec 7, 2016

### Ray Vickson

No, they differ by $\pi/2$ radians. It is a great mistake to mix trigonometric integrals with degree measurements; the calculus of trig functions is in radian units only!

4. Dec 7, 2016

5. Dec 7, 2016

### Any Help

then what's the integral that gives arccos u?

6. Dec 7, 2016

### Ray Vickson

There are two antiderivatives of $1/\sqrt{1-x^2}$; they are $\arcsin(x)+C$ and $-\arccos(x)+C$. Typically, sources like tables of integrals, or computer algebra packages, give only ONE of the two, and those seem to always be the $\arcsin(x)$ version. When you do it manually you are certainly allowed to give $-\arccos(x) + C$ as an answer; it is absolutely correct, because it is the same as $\arcsin(x) + K$ for some constant $K$ that is different from the constant $C$.