We start by using the techniques we would use to find the indefinite integral. Use integration by parts.
Let #u = t^2# and #dv = costdt#. Then #du = 2tdt# and #v = sint#.
By integration by parts:
#int(t^2cost)dt = sint(t^2) - int(sint(2t))#
Use integration by parts again for the remaining integral.
#int(sint(2t))dt = -cost(t) - int(-cost(2))#
#int(sint(2t))dt = -2tcos(t) - 2int(-cost)#
#int(sint(2t))dt = -2tcos(t) + 2sint + C#
Resubstitute:
#int(t^2cost)dt = t^2sint - (-2tcos(t) + 2sint) + C#
#int(t^2cost) = t^2sint + 2tcos(t) - 2sint + C#
#int(t^2cost) = (t^2- 2)sint + 2tcost + C#
Since this is a definite integral, we can forget about the "#C#".
#int_(0)^(pi/2)(t^2cost) = ((pi/2)^2 - 2)sin(pi/2) + 2(pi/2)cos(pi/2) - ((0^2 - 2)sin(0) + 2(0)(cos(0))#
#int_0^(pi/2)(t^2cost) = (pi^2/4 - 2)(1) + pi(0) - (-2(0) + 0 + 0)#
#int_0^(pi/2)(t^2cost) ~=0.4674#
Hopefully this helps!
