Consider the integral arccos(3y)19y2 dy\displaystyle -\int\frac{{{\arccos{{\left({3}{y}\right)}}}}}{{\sqrt{{{1}-{9}{y}^{{2}}}}}}\ {\left.{d}{y}\right.}:

a) This can be transformed into a basic integral by letting

u=\displaystyle {u}=   and

du=\displaystyle {d}{u}=  

b) After performing the substitution, you obtain the integral (in terms of u)

\displaystyle \int  

c) Once we integrate and substitute, the final answer in terms of y is: