How do you find the antiderivative of #cos^(-1)x dx#?

1 Answer
Jun 18, 2017

Use parts, or substitution and then parts.

Explanation:

#int cos^-1 x dx#

Let #u = cos^-1x# so #du = -1/(sqrt(1-x^2) dx# and

let #dv = dx#, so #v = x#

#uv-int vdu = xcos^-1x + int x/sqrt(1-x^2) dx#

Use substitution #u = 1-x^2# to finish with

#int cos^-1x dx = xcos^-1 x - sqrt(1-x^2) +C#

OR replace the inverse cosine first

Let #theta = cos^-1x# so that #cos theta = x# and -sin theta d theta = dx#

The integral becomes

#-inttheta sin theta d theta# Which may be integrated by parts with #u = theta# and #dv = sin theta d theta#

When finished we get

#- (sin theta - theta cos theta) + C#

And reversing the substitution gets #xcos^-1 x - sqrt(1-x^2) +C# as above.