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

Question

#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

889 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-09T08:25:11

See below

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 :)