Solution for sin6xcos4x dx?
I assume you want the integral?
Using your product-to-sum formula,
recall that sin6x cos4x = 1/2 (sin2x + sin10x)
Now the integral is easy.
To solve the integral of sin(6x)cos(4x) dx, we can use the double-angle formula for trigonometric functions, specifically the formula:
sin(A + B) = sinAcosB + cosAsinB
In this case, we have sin(6x)cos(4x), so we can rewrite it as:
2sin((6x + 4x) / 2)cos((6x - 4x) / 2)
Simplifying further, we get:
2sin(5x)cos(x)
Now, let's use the product-to-sum formula for trigonometric functions, which states:
sinAcosB = (1/2)(sin(A + B) + sin(A - B))
Applying this formula to our expression, we have:
2sin(5x)cos(x) = sin(5x + x) + sin(5x - x)
These can be further simplified as follows:
sin(5x + x) = sin(6x)
sin(5x - x) = sin(4x)
Therefore, the integral of sin(6x)cos(4x) dx can be written as:
∫(sin(6x)cos(4x)) dx = ∫(2sin(5x)cos(x)) dx
Simplifying further:
= ∫(sin(6x)) dx + ∫(sin(4x)) dx
Now, we can integrate each term separately:
Integrating sin(6x):
= -cos(6x) / 6
Integrating sin(4x):
= -cos(4x) / 4
Therefore, the solution to the integral ∫(sin(6x)cos(4x)) dx is:
= -cos(6x) / 6 - cos(4x) / 4 + C
where C is the constant of integration.