# intcos^2xtan^3xdx = intcos^2x(sin^3x/cos^3x)dx #
# :. intcos^2xtan^3xdx = intsin^3x/cosxdx #
We can perform this integral by a simple try substitution:
Let # u=cosx => (du)/dx=-sinx #, so #-int ... du=int...sinxdx#
# :. intcos^2xtan^3xdx = intsin^2xsinx/cosxdx #
# :. intcos^2xtan^3xdx = int(1-cos^2x)sinx/cosxdx #
Then if we perform the substitution we get:
# intcos^2xtan^3xdx = int(1-u^2)(-1/u)du #
# :. intcos^2xtan^3xdx = int(u^2-1)/udu #
# :. intcos^2xtan^3xdx = intu-1/udu #
# :. intcos^2xtan^3xdx = 1/2u^2-ln|u| + C#
# :. intcos^2xtan^3xdx = 1/2cosx^2-ln|cosx| + C#
