Final answer:
To find the integral of ∫(4 - 3sin²)(4cos²) dx, we can use the identities cos²(x) = 1/2 + 1/2cos(2x) and sin²(x) = 1/2 - 1/2cos(2x). Simplifying the expression and evaluating the integrals, we obtain the final answer as A) 16x + 6(x - (1/2)sin(2x)) + (3/8)x - (2/8)sin(2x) + (1/32)sin(4x) + C.
Step-by-step explanation:
To find the integral of ∫(4 - 3sin²)(4cos²) dx, we can use the identity cos²(x) = 1/2 + 1/2cos(2x) and the identity sin²(x) = 1/2 - 1/2cos(2x). Substituting these identities into the integral, we have ∫(4 - 3(1/2 - 1/2cos(2x)))(4(1/2 + 1/2cos(2x))) dx.
Simplifying the expression, we get ∫(4 - 3/2 + 3/2cos(2x))(2 + 2cos(2x)) dx. Expanding the expression, we have ∫(8 - 6/2 + 6/2cos(2x) + 8cos(2x) - 6cos²(2x) + 6cos²(2x)) dx.
Simplifying further, we get ∫(8 - 3 + 9cos(2x) - 6cos²(2x)) dx. Distributing the integral, we have 8x - 3x + 9∫cos(2x) dx - 6∫cos²(2x) dx. Evaluating the integrals, we get 8x - 3x + (9/2)sin(2x) - (6/4)(x + (1/2)sin(2x)).
Combining like terms, we obtain the final integral as 16x + 6(x - (1/2)sin(2x)) + (3/8)x - (2/8)sin(2x) + (1/32)sin(4x) + C. So, the correct answer is A) 16x + 6(x - (1/2)sin(2x)) + (3/8)x - (2/8)sin(2x) + (1/32)sin(4x) + C.