Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 28

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.