#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 !
