int x^3 dx - int csc(4x) dx∫x3dx−∫csc(4x)dx
1/4 x^4 - int 1/sin(4x) dx14x4−∫1sin(4x)dx
The table of integration tells us that int 1/sin(x) = -ln(cot(x) + csc(x))∫1sin(x)=−ln(cot(x)+csc(x))
Using a uu substitution, let u=4xu=4x. Then du = 4 dxdu=4dx and 1/4 du = dx14du=dx
1/4 int 1/sin(u) du = -1/4(ln(cot(u) - csc(u))) + C14∫1sin(u)du=−14(ln(cot(u)−csc(u)))+C
Plugging back in gives
f(x) = 1/4 x^4 + 1/4(ln(cot(4x) - csc(4x))) + Cf(x)=14x4+14(ln(cot(4x)−csc(4x)))+C
Given the condition that f(pi/12)=-1f(π12)=−1, we can determine CC.
1/4 (pi/12)^4 + 1/4(ln(cot(4 pi/12) - csc(4 pi/12))) + C=-114(π12)4+14(ln(cot(4π12)−csc(4π12)))+C=−1
pi^4/82944 + 1/4(ln(1/sqrt(3) - 2/sqrt(3))) + C=-1π482944+14(ln(1√3−2√3))+C=−1
Given how ugly this is, it makes sense to plug it into a calculator.
-0.1362 + C ~~ -1−0.1362+C≈−1
C ~~ -1.1362C≈−1.1362
Thus,
f(x) ~~ 1/4 x^4 + 1/4(ln(cot(4x) - csc(4x))) -1.1362f(x)≈14x4+14(ln(cot(4x)−csc(4x)))−1.1362
