Evaluate the integral
interval from [0 to pi] t sin(3t)dt
Use integration by parts
u=t and dv=sin(3t)dt. then du=dt and v=-cos(3t)/3
here is my problem but Im having problem to solve with pi.
∫t sin(3t)dt
= -tcos(3t)/3 - ∫[-cos(3t)/3]dt
=-tcos(3t)/3 + sin(3t)/9 ](o,pi)
Your integral is correct, so we get
[ -πcos(3π) /3 + sin(3π) /9] - [ 0 + sin(0)/9]
= -π(-1)/3 + 0/9 - 0
= π/3
its correct answer. thanks
To evaluate the given integral using integration by parts, we start by choosing u = t and dv = sin(3t)dt
Differentiating u gives du = dt and integrating dv gives v = -cos(3t)/3
Using the formula for integration by parts:
∫u dv = uv - ∫v du, we have
∫t sin(3t)dt = -t cos(3t)/3 - ∫(-cos(3t)/3) dt
Next, we can simplify the integral:
∫t sin(3t)dt = -t cos(3t)/3 + ∫(cos(3t)/3) dt
To evaluate the integral ∫(cos(3t)/3) dt, we integrate with respect to t:
∫(cos(3t)/3) dt = (∫cos(3t) dt)/3 = sin(3t)/9 + C
Now we can substitute the values into the expression:
∫t sin(3t)dt = -t cos(3t)/3 + sin(3t)/9 + C
To evaluate the definite integral over the interval [0, pi], we substitute the upper and lower limits of integration into the expression:
∫[0 to pi] t sin(3t)dt = [-pi cos(3pi)/3 + sin(3pi)/9] - [0 cos(0)/3 + sin(0)/9]
= [-pi cos(3pi)/3 + sin(3pi)/9] - [0 + 0]
= -pi cos(3pi)/3 + sin(3pi)/9
Since cos(3pi) = cos(pi), which equals -1, and sin(3pi) = sin(pi), which equals 0, we have:
-pi cos(3pi)/3 + sin(3pi)/9 = -(-pi)/3 = pi/3
Therefore, the value of the definite integral over the interval [0, pi] is pi/3.
To evaluate the integral ∫(0 to π) t sin(3t) dt using integration by parts, we start by choosing u and dv:
u = t
dv = sin(3t) dt
Taking the derivatives of u and finding the antiderivative of dv:
du = dt
v = -cos(3t)/3
Now we can use the formula for integration by parts:
∫(u dv) = uv - ∫(v du)
Plugging in the values:
∫(0 to π) t sin(3t) dt = (t * (-cos(3t)/3)) - ∫(-cos(3t)/3) dt
Now let's simplify this expression:
= -t * cos(3t)/3 + ∫cos(3t)/3 dt
Integrating ∫cos(3t)/3 dt:
= -t * cos(3t)/3 + sin(3t)/9
Now we can evaluate this expression at the limits [0, π]:
= [-π * cos(3π)/3 + sin(3π)/9] - [0 * cos(0)/3 + sin(0)/9]
Since cos(3π) = cos(π) = -1 and sin(3π) = sin(π) = 0, and cos(0) = 1 and sin(0) = 0:
= [(-π * -1)/3 + 0/9] - [0 + 0/9]
= π/3
So the value of the integral ∫(0 to π) t sin(3t) dt is π/3.