how to find the integral of (3x+pi)cos3xdx from [pi over 6, pi over 3}
integrate by parts
int u dv = u v - int v du
dv = cos 3 x dx
so
v = (1/3) sin 3x
u = 3 x + pi
du = 3 dx
so
u v - (3/3)integral of sin 3x dx
put in your limits
(3x+pi)(sin 3x)- int sin3x dx
at pi/3 minus at pi/6
i really appreciate the work it helps me a lot thank you
To find the integral of the given function, (3x + π)cos(3x)dx, over the interval [π/6, π/3], you can follow these steps:
Step 1: Applying the integration rules, let's find the integral of cos(3x) first.
∫cos(3x) dx = (1/3)sin(3x) + C --- (1)
(where C is the constant of integration)
Step 2: Now, we need to integrate the term (3x + π).
First, integrate the term 3x:
∫3x dx = (3/2)x^2 + C1 --- (2)
(where C1 is the constant of integration)
Next, integrate the constant term π:
∫π dx = πx + C2 --- (3)
(where C2 is the constant of integration)
Combining (2) and (3), we get:
∫(3x + π) dx = (3/2)x^2 + πx + C3 --- (4)
(where C3 = C1 + C2, another constant of integration)
Step 3: Now, multiply the result from Step 2 with the initial integration from Step 1:
∫(3x + π)cos(3x) dx = ∫(3x + π) dx * ∫cos(3x) dx
= [(3/2)x^2 + πx + C3] * [(1/3)sin(3x) + C]
Step 4: Apply the limits to the integral expression obtained in Step 3.
∫(3x + π)cos(3x) dx = [(3/2)(π/3)^2 + π(π/3) + C3] * [(1/3)sin(3(π/3))]
- [(3/2)(π/6)^2 + π(π/6) + C3] * [(1/3)sin(3(π/6))]
Simplifying further, we obtain the final numerical result.
Note: The final result will involve calculations and simplification based on the numerical values of π/3 and π/6, as well as the constants of integration (C, C1, C2, C3).