int cos6xcos4xdx =
Use cos(A)cos(B)=(1/2)(cos(A+B)+cos(A-B))
∫ cos(6x) cos(4x) dx
= (1/2)∫ cos(10x)+cos(2x) dx
= (1/2)((1/10)sin(10x)+(1/2)sin(2x)) + C
= (1/20)sin(10x)+(1/4)sin(2z) + C
To evaluate the integral ∫ cos(6x)cos(4x)dx, we'll need to use the product-to-sum formula for cosine, which states:
cos(A)cos(B) = (1/2)(cos(A - B) + cos(A + B))
Using this formula, we can rewrite the integral as:
∫ cos(6x)cos(4x)dx = (1/2) ∫ (cos(6x - 4x) + cos(6x + 4x)) dx
Simplifying the expression inside the integral:
= (1/2) ∫ (cos(2x) + cos(10x)) dx
Now we can integrate term by term:
= (1/2) ∫ cos(2x) dx + (1/2) ∫ cos(10x) dx
Integrating each term:
= (1/2) [ (1/2)sin(2x) ] + (1/20) [ (1/10)sin(10x) ]
= (1/4)sin(2x) + (1/200)sin(10x)
Therefore, the indefinite integral of cos(6x)cos(4x) with respect to x is:
(1/4)sin(2x) + (1/200)sin(10x) + C
Where C is the constant of integration.