Question

Integral from 0 to 2pi of t²sin(2t)

Answer 1

To evaluate the integral from 0 to 2pi of t^2sin(2t), we can use integration by parts. The result is (-1/2 t^2 cos(2t) + 1/2 (2t sin(2t) - 2 cos(2t))) + C

Step-by-step explanation:

To evaluate the integral ∫ from 0 to 2π of t²sin(2t), we can use integration by parts. Let u = t² and dv = sin(2t) dt. Differentiating u gives du = 2t dt, and integrating dv gives v = -1/2 cos(2t). Now we can use the formula for integration by parts: ∫ u dv = uv - ∫ v du. Plugging in the values, we get:

∫ t²sin(2t) dt = -1/2 t² cos(2t) − ∫ -1/2 cos(2t) 2t dt

Simplifying further, we have:

∫ t²sin(2t) dt = -1/2 t² cos(2t) + 1/2 ∫ cos(2t) 2t dt

Now we can integrate the remaining term by parts again, with u = 2t and dv = cos(2t) dt. After differentiating u and integrating dv, we get:

∫ t²sin(2t) dt = -1/2 t² cos(2t) + 1/2 (2t sin(2t) - ∫ 2 sin(2t) dt)

Finally, integrating the remaining term, we have:

∫ t²sin(2t) dt = -1/2 t² cos(2t) + 1/2 (2t sin(2t) + −2 cos(2t)) + C