Question Detail

To integrate ∫sin(4x)cos(2x) dx, use the product-to-sum trigonometric identity: sin A cos B = (1/2)[sin(A+B) + sin(A-B)].Here, A = 4x and B = 2x.sin(4x)cos(2x) = (1/2)[sin(4x+2x) + sin(4x-2x)]= (1/2)[sin(6x) + sin(2x)]Now, integrate the expression:∫(1/2)[sin(6x) + sin(2x)] dx = (1/2)∫sin(6x) dx + (1/2)∫sin(2x) dx= (1/2)(-cos(6x)/6) + (1/2)(-cos(2x)/2) + C= -cos(6x)/12 - cos(2x)/4 + C= -1/12 (cos(6x) + 3cos(2x)) + CComparing this with option C: 1/12 (cos 6x + 3 cos 2x), there is a sign difference. The mathematically correct answer should be negative.