Evaluate the Integral

Problem

(∫_0^2*π)(t2*sin(2*t)*d(t))

Solution

1. Identify the method of integration by parts, which states (∫_^)(u*d(v))=u*v−(∫_^)(v*d(u)) Let u=t2 and d(v)=sin(2*t)*d(t)

2. Calculate the differentials and integrals for the first application: d(u)=2*t*d(t) and v=−1/2*cos(2*t)

3. Apply the integration by parts formula:

(∫_^)(t2*sin(2*t)*d(t))=−(t2)/2*cos(2*t)−(∫_^)(−t*cos(2*t)*d(t))

(∫_^)(t2*sin(2*t)*d(t))=−(t2)/2*cos(2*t)+(∫_^)(t*cos(2*t)*d(t))

4. Apply integration by parts a second time for the remaining integral (∫_^)(t*cos(2*t)*d(t)) Let u=t and d(v)=cos(2*t)*d(t) which gives d(u)=d(t) and v=1/2*sin(2*t)

5. Substitute the result of the second integration by parts back into the expression:

(∫_^)(t*cos(2*t)*d(t))=t/2*sin(2*t)−(∫_^)(1/2*sin(2*t)*d(t))

(∫_^)(t*cos(2*t)*d(t))=t/2*sin(2*t)+1/4*cos(2*t)

6. Combine all terms to find the general antiderivative:

(∫_^)(t2*sin(2*t)*d(t))=−(t2)/2*cos(2*t)+t/2*sin(2*t)+1/4*cos(2*t)

7. Evaluate the definite integral from 0 to 2*π by plugging in the upper and lower limits:

[−(t2)/2*cos(2*t)+t/2*sin(2*t)+1/4*cos(2*t)](2*π)0

8. Simplify the evaluation at t=2*π

−((2*π)2)/2*cos(4*π)+(2*π)/2*sin(4*π)+1/4*cos(4*π)

−2*π2*(1)+π*(0)+1/4*(1)=−2*π2+1/4

9. Simplify the evaluation at t=0

−0/2*cos(0)+0/2*sin(0)+1/4*cos(0)

0+0+1/4*(1)=1/4

10. Subtract the lower limit value from the upper limit value:

(−2*π2+1/4)−(1/4)=−2*π2

Final Answer

(∫_0^2*π)(t2*sin(2*t)*d(t))=−2*π2

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