What is the result of int_0^(pi/2) t^2costdt?

1 Answer
Dec 13, 2016

int_0^(pi/2)t^2cost ~= 0.4674

Explanation:

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!