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

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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.`

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