If `int _(0)^(2)(3x ^(2) -3x +1) cos (x ^(3) -3x ^(2)+4x -2) dx = a sin (b),` where a and b are positive integers. Find the value of `(a+b).`

Answer

To solve the integral

\[ I = \int_{0}^{2} (3x^2 - 3x + 1) \cos(x^3 - 3x^2 + 4x - 2) \, dx, \]

we can use the property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = 2 \). Thus, we have:

\[ I = \int_{0}^{2} (3(2-x)^2 - 3(2-x) + 1) \cos((2-x)^3 - 3(2-x)^2 + 4(2-x) - 2) \, dx. \]

Now, we simplify the expression \( 3(2-x)^2 - 3(2-x) + 1 \):

1. Calculate \( (2-x)^2 \):
\[ (2-x)^2 = 4 - 4x + x^2. \]
Thus,
\[ 3(2-x)^2 = 3(4 - 4x + x^2) = 12 - 12x + 3x^2. \]

2. Calculate \( -3(2-x) \):
\[ -3(2-x) = -6 + 3x. \]

3. Combine these results:
\[ 3(2-x)^2 - 3(2-x) + 1 = (12 - 12x + 3x^2) + (-6 + 3x) + 1 = 3x^2 - 9x + 7. \]

Next, we simplify the argument of the cosine function:

1. Calculate \( (2-x)^3 \):
\[ (2-x)^3 = 8 - 12x + 6x^2 - x^3. \]

2. Calculate \( -3(2-x)^2 \):
\[ -3(2-x)^2 = -3(4 - 4x + x^2) = -12 + 12x - 3x^2. \]

3. Calculate \( 4(2-x) \):
\[ 4(2-x) = 8 - 4x. \]

4. Combine these results:
\[ (2-x)^3 - 3(2-x)^2 + 4(2-x) - 2 = (8 - 12x + 6x^2 - x^3) + (-12 + 12x - 3x^2) + (8 - 4x) - 2. \]
Simplifying this gives:
\[ -x^3 + 3x^2 - 4x + 2. \]

Now, we can express the integral \( I \) as:

\[ I = \int_{0}^{2} (3x^2 - 3x + 1) \cos(x^3 - 3x^2 + 4x - 2) \, dx + \int_{0}^{2} (3x^2 - 9x + 7) \cos(-x^3 + 3x^2 - 4x + 2) \, dx. \]

This leads to:

\[ 2I = \int_{0}^{2} \left[ (3x^2 - 3x + 1) + (3x^2 - 9x + 7) \right] \cos(x^3 - 3x^2 + 4x - 2) \, dx. \]

Combining the terms gives:

\[ 2I = \int_{0}^{2} (6x^2 - 12x + 8) \cos(x^3 - 3x^2 + 4x - 2) \, dx. \]

Now, we can factor out the constant:

\[ 2I = 2 \int_{0}^{2} (3x^2 - 6x + 4) \cos(x^3 - 3x^2 + 4x - 2) \, dx. \]

Thus, we have:

\[ I = \int_{0}^{2} (3x^2 - 6x + 4) \cos(x^3 - 3x^2 + 4x - 2) \, dx. \]

Next, we perform a substitution. Let:

\[ t = x^3 - 3x^2 + 4x - 2. \]

Then, we find \( dt \):

\[ dt = (3x^2 - 6x + 4) \, dx. \]

Now we need to evaluate the limits when \( x = 0 \) and \( x = 2 \):

- When \( x = 0 \):
\[ t = 0^3 - 3(0)^2 + 4(0) - 2 = -2. \]

- When \( x = 2 \):
\[ t = 2^3 - 3(2^2) + 4(2) - 2 = 8 - 12 + 8 - 2 = 2. \]

Thus, the integral becomes:

\[ I = \int_{-2}^{2} \cos(t) \, dt. \]

Now, we evaluate this integral:

\[ I = \left[ \sin(t) \right]_{-2}^{2} = \sin(2) - \sin(-2) = \sin(2) + \sin(2) = 2\sin(2). \]

Thus, we have:

\[ I = 2 \sin(2). \]

From the problem statement, we have \( I = a \sin(b) \) where \( a = 2 \) and \( b = 2 \).

Finally, we find \( a + b \):

\[ a + b = 2 + 2 = 4. \]

Therefore, the answer is:

\[ \boxed{4}. \]