Integrate function [closed]

Question

I want to find the constant in front of the term of the integral that only contains $x$ in the following function: $$f(x) = \frac{(56-15 \sin x + \sin^2 x)^2}{6-5 \sin x + \sin^2 x}$$

This is my input in Mathematica:

Integrate[((56 - 15 Sin[x] + Sin[x]^2*x)^2)/(6 - 5 Sin[x] + Sin[x]^2), x] 

And this is the output:

1/2 x (450 - 38 x + 13 x^2) - 5 (-2 - 6 x + x^2) Cos[x] - 
 1/4 x Cos[2 x] - (1/Sqrt[2])
 I (1/2 (11 + 9 x)^2 Log[1 + I (3 - 2 Sqrt[2]) E^(I x)] - 
    1/2 (11 + 9 x)^2 Log[1 + I (3 + 2 Sqrt[2]) E^(I x)] - 
    9 I (11 + 9 x) PolyLog[2, I (-3 + 2 Sqrt[2]) E^(I x)] + 
    9 I (11 + 9 x) PolyLog[2, -I (3 + 2 Sqrt[2]) E^(I x)] + 
    81 PolyLog[3, I (-3 + 2 Sqrt[2]) E^(I x)] - 
    81 PolyLog[3, -I (3 + 2 Sqrt[2]) E^(I x)]) - (1/Sqrt[3])
 4 I (-(13 + 2 x)^2 Log[(2 + Sqrt[3] + I E^(I x))/(
      2 + Sqrt[3])] + (13 + 2 x)^2 Log[
      1 - (I E^(I x))/(-2 + Sqrt[3])] - 
    4 I (13 + 2 x) PolyLog[2, (I E^(I x))/(-2 + Sqrt[3])] + 
    4 I (13 + 2 x) PolyLog[2, -((I E^(I x))/(2 + Sqrt[3]))] + 
    8 PolyLog[3, (I E^(I x))/(-2 + Sqrt[3])] - 
    8 PolyLog[3, -((I E^(I x))/(2 + Sqrt[3]))]) + 
 10 (-3 + x) Sin[x] - 1/8 (-1 + 2 x^2) Sin[2 x]

Therefore, the constant before the $x$-term is $450/2=225$, however that is incorrect. What am I doing wrong?

Answer 1

If TeX formula is the correct one(see comment @b.gates.you.know.what ) you get

int=Integrate[((56 - 15 Sin[x] + Sin[x]^2 )^2)/(6 - 5 Sin[x] + Sin[x]^2),x]

x- coefficient evaluates to

Coefficient[int,x] (*413/2)*)

Answer 2

Why do you think this is incorrect?

Consider:

f = ((56 - 15 Sin[x] + Sin[x]^2*x)^2)/(6 - 5 Sin[x] + Sin[x]^2);
int = Integrate[f, x]

Now if you take the derivative:

D[int, x] == f // Simplify

True

Seems to be o.k.