`int_0^(2pi) t^2 sin(2t) dt` Evaluate the integral

You need to
use the integration by parts for `int_0^(2pi) t^2*sin(2t)dt`   such that:

`int udv = uv - int vdu`

`u = t^2 => du = 2tdt`

`dv = sin 2t=> v =(-cos 2t)/2`

`int_0^(2pi) t^2*sin(2t)dt =
t^2*(-cos 2t)/2|_0^(2pi) + int_0^(2pi) t*cos 2t dt`

You need to use the
integration by parts for `int_0^(2pi) t*cos 2t dt`  such that:

`u = t=> du
= dt`

`dv = cos 2t=> v = (sin 2t)/2`

`int_0^(2pi) t*cos
2t dt = t*(sin 2t)/2|_0^(2pi) - (1/2)int_0^(2pi) sin 2t dt`

`int_0^(2pi)
t*cos 2t dt = t*(sin 2t)/2|_0^(2pi) +  (cos 2t)/4|_0^(2pi) `

`int_0^(2pi)
t^2*sin(2t)dt = t^2*(-cos 2t)/2|_0^(2pi) + t*(sin 2t)/2|_0^(2pi) +  (cos 2t)/4|_0^(2pi)`
 

Using the fundamental theorem of calculus yields:

`int_0^(2pi) t^2*sin(2t)dt = (2pi)^2*(-cos 4pi)/2 + 0*(cos 0)/2  + 2pi*(sin 4pi)/2 -
0 +  (cos 4pi)/4 - (cos 0)/4`

`int_0^(2pi) t^2*sin(2t)dt = -2(pi)^2 + 1/4 -
1/4`

`int_0^(2pi) t^2*sin(2t)dt = -2(pi)^2 `

Hence, evaluating the integral, using  integration by parts,
yields `int_0^(2pi) t^2*sin(2t)dt = -2(pi)^2.`