What is the antiderivative of #(arccos(x/6))^2#?

1 Answer
Aug 6, 2018

#I=xcos^-1(x/6)^2-12cos^-1(x/6)sqrt(1-x^2/36)-2x+C#, #C in RR#

Explanation:

#I=intcos^(-1)(x/6)^2dx#

Let #X=x/6#

#dx=6dX#

So:

#I=6intcos^(-1)(X)^2dX#

Using intregration by parts :
#f'(X)=1#, #f(X)=X#,

#g(X)=cos^-1(X)^2#, #g'(X)=(-2cos^-1(X))/sqrt(1-X^2)#

So :

#I=6Xcos^(-1)(X)^2+12int(Xcos^(-1)(X))/sqrt(1-X^2)dX#

We will use another integration by parts,

#g(X)=cos^-1(X)#, #g'(X)=-1/sqrt(1-X^2)#

#f'(X)=X/sqrt(1-X^2)#, #f(X)=-sqrt(1-X^2)# (Here's a proof.)

So :

#I=6Xcos^(-1)(X)^2+12(-cos^-1(X)sqrt(1-X^2)-int(-sqrt(1-X^2))/(-sqrt(1-X^2))dX)#

#=6Xcos^(-1)(X)^2-12cos^-1(X)sqrt(1-X^2)-12int1dX#

#=6Xcos^(-1)(X)^2-12cos^-1(X)sqrt(1-X^2)-12X+C#, #C in RR#

Substitute back, and finally we have :

#I=xcos^-1(x/6)^2-12cos^-1(x/6)sqrt(1-x^2/36)-2x+C#, #C in RR#

\0/ Here's our answer !