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 a property of definite integrals. The property states that:

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

In our case, \( a = 0 \) and \( b = 2 \), so we can rewrite the integral as:

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

### Step 1: Substitute \( x \) with \( 2 - x \)

First, we will calculate the expression \( 3(2-x)^2 - 3(2-x) + 1 \):

\[ 3(2-x)^2 = 3(4 - 4x + x^2) = 12 - 12x + 3x^2 \]
\[ -3(2-x) = -6 + 3x \]
\[ 1 = 1 \]

Combining these, we have:

\[ 3(2-x)^2 - 3(2-x) + 1 = (12 - 12x + 3x^2) + (-6 + 3x) + 1 = 3x^2 - 9x + 7 \]

### Step 2: Calculate the cosine term

Next, we calculate the expression for the cosine term:

\[ (2-x)^3 - 3(2-x)^2 + 4(2-x) - 2 \]

Calculating each part:

\[ (2-x)^3 = 8 - 12x + 6x^2 - x^3 \]
\[ -3(2-x)^2 = -3(4 - 4x + x^2) = -12 + 12x - 3x^2 \]
\[ 4(2-x) = 8 - 4x \]
\[ -2 = -2 \]

Combining these gives:

\[ (8 - 12x + 6x^2 - x^3) + (-12 + 12x - 3x^2) + (8 - 4x) - 2 \]

Simplifying this:

\[ = 8 - 12 + 8 - 2 + (-x^3 + 6x^2 - 3x^2 - 4x + 12x) = -x^3 + 3x^2 + 0x + 2 \]

Thus, the cosine term becomes:

\[ \cos((2-x)^3 - 3(2-x)^2 + 4(2-x) - 2) = \cos(-x^3 + 3x^2 + 2) \]

### Step 3: Combine both results

Now we can rewrite the integral \( I \):

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

### Step 4: Add the two integrals

Adding the two integrals, we have:

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

This simplifies to:

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

### Step 5: Factor out the constant

Now we can factor out the 2:

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

### Step 6: Change of variable

Let \( t = x^3 - 3x^2 + 4x - 2 \). Then \( dt = (3x^2 - 6x + 4) \, dx \).

### Step 7: Evaluate the integral

The integral becomes:

\[ I = \int_{t(0)}^{t(2)} \cos(t) \, dt \]

Calculating the limits:

At \( x = 0 \):

\[ t(0) = 0^3 - 3(0)^2 + 4(0) - 2 = -2 \]

At \( x = 2 \):

\[ t(2) = 2^3 - 3(2)^2 + 4(2) - 2 = 8 - 12 + 8 - 2 = 2 \]

Thus,

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

### Final Result

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

Thus, \( a + b = 2 + 2 = 4 \).