If `I = int(2cos^(2)x-1)(4cos^(3)x-3cosx)dx=Asin5x+(1)/(2)sinx+C` then A is equal to

Answer

To solve the integral \( I = \int (2\cos^2 x - 1)(4\cos^3 x - 3\cos x) \, dx \) and find the value of \( A \) in the expression \( I = A \sin 5x + \frac{1}{2} \sin x + C \), we can follow these steps:

### Step 1: Simplify the integrand
We start by rewriting \( 2\cos^2 x - 1 \) using the double angle identity:
\[ 2\cos^2 x - 1 = \cos 2x \]
Thus, the integral becomes:
\[ I = \int \cos 2x (4\cos^3 x - 3\cos x) \, dx \]

### Step 2: Factor the expression
Next, we can factor \( 4\cos^3 x - 3\cos x \):
\[ 4\cos^3 x - 3\cos x = \cos x (4\cos^2 x - 3) \]
Using the identity \( 4\cos^2 x - 3 = \cos 3x \), we can rewrite the integrand:
\[ I = \int \cos 2x \cos x \cos 3x \, dx \]

### Step 3: Use the product-to-sum identities
We can apply the product-to-sum identities:
\[ \cos A \cos B = \frac{1}{2} (\cos(A + B) + \cos(A - B)) \]
Applying this to \( \cos 2x \cos 3x \):
\[ \cos 2x \cos 3x = \frac{1}{2} (\cos(5x) + \cos(-x)) = \frac{1}{2} (\cos 5x + \cos x) \]
Thus, we have:
\[ I = \int \frac{1}{2} \cos x \left( \frac{1}{2} (\cos 5x + \cos x) \right) \, dx \]
This simplifies to:
\[ I = \frac{1}{4} \int \cos x \cos 5x \, dx + \frac{1}{4} \int \cos^2 x \, dx \]

### Step 4: Integrate each term
1. For \( \int \cos x \cos 5x \, dx \):
\[ \int \cos x \cos 5x \, dx = \frac{1}{2} \left( \int \cos(6x) \, dx + \int \cos(4x) \, dx \right) \]
This results in:
\[ = \frac{1}{2} \left( \frac{\sin 6x}{6} + \frac{\sin 4x}{4} \right) \]

2. For \( \int \cos^2 x \, dx \):
\[ \int \cos^2 x \, dx = \frac{1}{2} \left( x + \frac{\sin 2x}{2} \right) \]

### Step 5: Combine results
Combining these results, we can express \( I \) as:
\[ I = \frac{1}{4} \left( \frac{1}{12} \sin 6x + \frac{1}{8} \sin 4x \right) + \frac{1}{8} \left( x + \frac{\sin 2x}{2} \right) + C \]

### Step 6: Compare coefficients
From the original expression \( I = A \sin 5x + \frac{1}{2} \sin x + C \), we can see that the coefficient of \( \sin 5x \) is \( \frac{1}{10} \) when we integrate \( \cos 5x \). Therefore, we conclude that:
\[ A = \frac{1}{10} \]

### Final Answer
Thus, the value of \( A \) is:
\[ \boxed{\frac{1}{10}} \]