To solve this we must recall that #int sin (u)du = -cos u +c#, where #u# is some function of x. In this case our u is #x^3#, so #du = 3x^2#. Thus our first part of the integral will yield #-cos (x^3)/3 +c#; double checking this by differentiation, we do indeed receive #3x^2sinx^3/3 = x^2sinx^3#.
For our second part, recall that #cot u = cos u/sin u#. Recall also that #d/(du)sin u = (cos u) du#. If our u is 2x, then du is 2. Recall further that #d/(du) ln u = (du)/u#. Using these facts...
#int -cot2xdx = int -(cos2x)/(sin2x) = int -(du)/(2u) = -1/2 int (du)/u = -1/2 lnu + c = -(ln sin 2x)/2 +c#
Thus our general f (x) is:
#f (x) = -cosx^3/3 - ln (sin 2x)/2 + c#
If #f (pi/8) = -1#, then...
#f (pi/8) = -cos (pi^3/512)/3 -ln sin (pi/4)/2 +c = -1-> c = -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2 #
Thus...
#f (x)=-cosx^3/3 -ln sin (2x)/2 -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2#
This answer is somewhat lengthy yes, but note that # -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2# is in fact constant and can be calculated; this calculation is left to the student as many professors would simply accept the current form.
