integrate 4xcos4x dx
We can use integration by parts for this problem:
Let u = 4x and dv = cos(4x) dx, then du/dx = 4 and v = (1/4)sin(4x)
Using the formula for integration by parts, we have:
∫4xcos(4x) dx = (4x)(1/4)sin(4x) - ∫(1/4)sin(4x)(4)dx
Simplifying, we get:
∫4xcos(4x) dx = x sin(4x) - (1/4)∫sin(4x) dx
Integrating the sine term, we have:
∫4xcos(4x) dx = x sin(4x) + (1/16)cos(4x) + C
Therefore, the solution is:
∫4xcos(4x) dx = x sin(4x) + (1/16)cos(4x) + C
To integrate 4x cos(4x) dx, you can use integration by parts. The formula for integration by parts is ∫ u dv = uv - ∫ v du.
Let's choose u = 4x and dv = cos(4x) dx. Then, we can calculate du and v.
Differentiating u = 4x:
du = 4 dx
Integrating dv = cos(4x) dx:
∫ dv = ∫ cos(4x) dx = (1/4) sin(4x)
Now we can plug in the values into the integration by parts formula:
∫ 4x cos(4x) dx = u ∫ dv - ∫ v du
= 4x * (1/4) sin(4x) - ∫ (1/4) sin(4x) * 4 dx
Simplifying further:
∫ 4x cos(4x) dx = x sin(4x) - (1/4) ∫ sin(4x) dx
= x sin(4x) + (1/16) cos(4x) + C
So, the final result after integrating 4x cos(4x) dx is x sin(4x) + (1/16) cos(4x) + C, where C is the constant of integration.