What is the integral of t^2sin (t)dt?

Integration by parts : ∫u dv = u v - ∫v du

∫t² sin(t) dt
= ∫t² d[-cos(t)]
= t² [-cos(t)] - ∫[-cos(t)] dt²
= -t² cos(t) + ∫cos(t) (2t dt)
= -t² cos(t) + 2 ∫t cos(t) dt
= -t² cos(t) + 2 ∫t d[sin(t)]
= -t² cos(t) + 2 [t sin(t) - ∫sin(t) dt]
= -t² cos(t) + 2 t sin(t) - 2 ∫sin(t) dt
= -t² cos(t) + 2 t sin(t) - 2 [-cos(t)] + C
= -t² cos(t) + 2 t sin(t) + 2 cos(t) + C

Integrate by parts, twice.

If you forget how, see this
http://www.mathsisfun.com/calculus/integration-by-parts.html

You should get
2t sin(t) - (t^2 - 2)cos(t)

let y = sin at
dy/da= t cosa t,
d^2y/da^2= -t^2 sin at,,
integrate y = - int t^2 sin at= -1/a cos at. Now you find d^2/da^2 ( -1/a cos at),
Negative sign goes out. Now take twice differential. You will get the answer.