#int x^3 dx - int csc(4x) dx#
#1/4 x^4 - int 1/sin(4x) dx#
The table of integration tells us that #int 1/sin(x) = -ln(cot(x) + csc(x))#
Using a #u# substitution, let #u=4x#. Then #du = 4 dx# and #1/4 du = dx#
#1/4 int 1/sin(u) du = -1/4(ln(cot(u) - csc(u))) + C#
Plugging back in gives
#f(x) = 1/4 x^4 + 1/4(ln(cot(4x) - csc(4x))) + C#
Given the condition that #f(pi/12)=-1#, we can determine #C#.
#1/4 (pi/12)^4 + 1/4(ln(cot(4 pi/12) - csc(4 pi/12))) + C=-1#
#pi^4/82944 + 1/4(ln(1/sqrt(3) - 2/sqrt(3))) + C=-1#
Given how ugly this is, it makes sense to plug it into a calculator.
#-0.1362 + C ~~ -1#
#C ~~ -1.1362#
Thus,
#f(x) ~~ 1/4 x^4 + 1/4(ln(cot(4x) - csc(4x))) -1.1362#
