# Trigonometric functions and integrals

## Homework Statement

I'm searching for the integral that gives arcosu

## Homework 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

## 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?

cnh1995
Homework Helper
Gold Member

## Homework Statement

I'm searching for the integral that gives arcosu

## Homework 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

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

Ray Vickson
Homework Helper
Dearly Missed
cos-1u= -sin-1u+90°.
Sine and cosine functions have a phase difference of 90°.

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!

cnh1995
Mark44
Mentor

No, they differ by π/2π/2\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!
then what's the integral that gives arccos u?

Ray Vickson