Evaluate integral by using integration by parts: intcos^-1x dx?

intcos^-1x dx

I'm stuck on one particular part, here's what I've got so far:

u=cos^-1x
du=-1/sqrt(1-x^2)dx
v=x
dv=dx

intudv=uv-intvdu
=xcos^-1x-int-x/sqrt(1-x^2)dx
t=1-x^2
dt=-2xdx
=xcos^-1x-int-x/sqrtt(dt/(-2x))
=xcos^-1x-intdt/(2sqrtt)

after that I don't know what to do

1 Answer
Apr 9, 2017

See below

Explanation:

For this bit that follows, I'd argue that the sub is a OTT and that, if you know the derivative of cos^(-1) x, you should see this pattern:

=-int-x/sqrt(1-x^2)dx =int x/sqrt(1-x^2)dx

color(red)(=int d/dx ( - sqrt(1-x^2) ) dx)

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

But moving forward with the sub:

intdt/(2sqrtt)

Set up for power rule if that helps visualise:

= 1/2 int t^(-1/2)dt

Apply power rule int z^n dz = z^(n+1)/(n+1) + C:

= 1/2 t^(1/2)/(1/2) + C

Reversing out of the sub:

= (1-x^2)^(1/2) + C

Overall, keep an eye on that minus sign :)