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

Question

113137 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T14:28:59

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

#int cos^(-1) x dx#

we know #d/dx ( cos^(-1) x ) = -1/sqrt(1-x^2)# so we can try set up an IBP and use that fact

#= int d/dx (x) cos^(-1) x dx#

#=x cos^(-1) x - int x d/dx ( cos^(-1) x ) dx#

#=x cos^(-1) x + int x*1/sqrt(1-x^2) dx#

we know that #d/dx (sqrt(1-x^2)) = 1/2 1/sqrt(1-x^2) (-2x) = -x/sqrt(1-x^2) #

#=x cos^(-1) x + int d/dx(-sqrt(1-x^2)) dx#

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