Expert: Ahmed Salami - 3/3/2006

Question
how do you integrate cos^(-1)x dx

Answer
Hi Errich,
Sorry for the time it took.
Did you mean the integral of arccos x?
Arccos x is the inverse cosine of x i.e if
x = cos y
y = arccos x
For our purpose, $ represents the integral sign.
We can integrate arccos x using the method of
integration by parts; which is
$u dv = uv - $v du
$arccos x dx = $(arccos x).1.dx
Taking u = arccos x and dv = 1.dx = dx
We now need to find du and v.
u = arccos x
x = cos u
dx/du = -sin u
du = dx/-sin u
but (cos u)^2 + (sin u)^2 = 1
(sin u)^2 = 1 - (cos u)^2
sin u = sqrt[1 - (cos u)^2]
     = sqrt(1 - x^2)     [remembering that x = cos u]
du = dx/sqrt(1 - x^2)
dv = dx
Integrating both sides
v = x
Making use of the integration by parts formula,
$arccos x dx = (arccos x).x - $xdx/sqrt(1 - x^2)
To find $xdx/sqrt(1 - x^2), we use the substitution
x = sin #
dx/d# = cos #
dx = cos# d#
$xdx/sqrt(1 - x^2) = $(sin# cos# d#)/sqrt[1 - (sin#)^2]
                  = $(sin# cos# d#)/sqrt[(cos#)^2]
                  = $(sin# cos# d#)/cos#
                  = $sin# d#
                  = -cos# + c   (c is a constant)
                  = -sqrt(1 - x^2) + c
Therefore, as a final answer
$arccos x dx = x(arccos x) - sqrt(1 - x^2) + c

If that is too elaborate, let's try another method.
Let u = arccos x
as before. Then we know that
x = cos u
dx/du = -sin u
dx = -sinu du
$(arccos x) dx becomes $u.(-sinu du) = -$u.sinu du
Using integration by parts and taking dv = sin u
v = -cos u
$u sinu du = u(-cos u) - $(-cos u).1.du
          = -u cosu + $cos u du
          = -u cosu + sinu + c
Since cosu = x, we know that sinu = sqrt(1 - x^2)
Substituting back into the equation,
$u sinu du = -(arccos x)x + sqrt(1 - x^2) + c
But $arccos x dx = -$u sinu du     (as shown above)
Therefore,
$arccos x dx = -$u sinu du
            = -[-(arccos x)x + sqrt(1 - x^2)] + c
$arccos x dx = x(arccos x) - sqrt(1 - x^2) + c
Just as we had before.

I hope i have helped. You can always get back to me.
Regards.