evaluate the integral.
from 0 to 4π
t^2 sin(2t) dt
To evaluate the integral ∫₀^(4π) t² sin(2t) dt, we can use integration by parts.
Let's choose u = t² and dv = sin(2t) dt.
Taking the derivatives and antiderivatives:
du = 2t dt (Differentiate u)
v = -1/2 cos(2t) (Integrate dv)
By using the integration by parts formula:
∫ u dv = uv - ∫ v du
Applying this formula, we have:
∫₀^(4π) t² sin(2t) dt = -1/2 t² cos(2t) ∣₀^(4π) - ∫₀^(4π) (-1/2 cos(2t)) (2t dt)
Now, let's calculate each part step by step:
1. Evaluating the first part:
-1/2 t² cos(2t) ∣₀^(4π)
= -1/2 (4π)² cos(2(4π)) - (-1/2 (0)² cos(2(0)))
= -1/2(16π²) cos(8π) - 0
As cos(8π) = cos(0) = 1:
= -8π²
2. Next, we can evaluate the second part:
∫₀^(4π) (-1/2 cos(2t)) (2t dt)
= -∫₀^(4π) t cos(2t) dt
To evaluate this integral, we can use integration by parts again.
Let's choose u = t and dv = cos(2t) dt.
Taking the derivatives and antiderivatives:
du = dt (Differentiate u)
v = 1/2 sin(2t) (Integrate dv)
By applying the integration by parts formula:
∫ t cos(2t) dt = (1/2 t sin(2t)) - ∫ (1/2 sin(2t)) dt
Simplifying this, we have:
∫₀^(4π) (-1/2 cos(2t)) (2t dt) = -(1/2 t sin(2t)) ∣₀^(4π) + ∫₀^(4π) (1/4 sin(2t)) dt
3. Evaluating the second part of the second integral:
∫₀^(4π) (1/4 sin(2t)) dt
= (1/4) ∫₀^(4π) sin(2t) dt
Since the integral of sin(2t) dt is -1/2 cos(2t) + C, we have:
(1/4) ∫₀^(4π) sin(2t) dt
= (1/4) (-1/2 cos(2t)) ∣₀^(4π)
= -1/8 cos(8π) + 1/8 cos(0)
= -1/8 - 1/8
= -1/4
Now, let's substitute these results back into the equation:
-∫₀^(4π) t cos(2t) dt = -(1/2 t sin(2t)) ∣₀^(4π) + ∫₀^(4π) (1/4 sin(2t)) dt
= -[(1/2 (4π) sin(2(4π))) - (1/2 (0) sin(2(0))) ] - 1/4
= -4π sin(8π) - 0 - 1/4
= -4π(0) - 0 - 1/4
= -1/4
Finally, putting it all together:
∫₀^(4π) t² sin(2t) dt = -1/2(8π²) - 1/4
= -4π² - 1/4
To evaluate the integral ∫(0 to 4π) t^2 sin(2t) dt, we can use integration by parts.
Integration by parts is a technique based on the product rule of differentiation. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
To use integration by parts, we need to choose u and dv. In this case, we can let:
u = t^2 (u is chosen such that its derivative is easier to handle)
dv = sin(2t) dt
Next, we need to find du and v. We differentiate u to find du, and integrate dv to find v.
Differentiating u = t^2:
du = 2t dt
Integrating dv = sin(2t) dt:
v = -1/2cos(2t) (integrate sin(2t) with respect to t)
Now, we can substitute these values into the integration by parts formula:
∫ t^2 sin(2t) dt = t^2 (-1/2cos(2t)) - ∫ (-1/2cos(2t)) (2t dt)
Simplifying this expression:
= -1/2t^2cos(2t) + 1/2 ∫ cos(2t) (2t dt)
We can further simplify the expression ∫ cos(2t) (2t dt) to complete the integration.
Finally, we plug the limits of integration (0 to 4π) into the expression and evaluate the integral.
you will need to use integration by parts, twice.
1st step:
u = t^2
dv=sin(2t) dt
du = 2t dt
v = -1/2 cos(2t)
That makes the first integration by parts: ?u dv = uv - ?v du
?t^2 sin(2t) dt
= -1/2 t^2 cos(2t) + ?t cos(2t) dt
Then repeat, letting
u = t
dv = cos(2t) dt
You should wind up with this:
http://www.wolframalpha.com/input/?i=%E2%88%AB+t%5E2+sin(2t)+dt
Then evaluate that at the limits.