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 !