Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 45

To evaluate the given integral problem: int_0^(pi/2) t^3 cos(t)dt , we may apply the First Fundamental Theorem of Calculus. It states that when continuous function f on closed interval [a,b] and F as indefinite integral of f then int_a^b f(x) dx = F(x)|_a^b or F(b)-F(a) .
The given integrand is f(t) =t^3 cos(t)dt on closed interval [0, pi/2] . To determine the indefinite integral F(t) , we may consider the formula from integration table. The integral int_0^(pi/2) t^3 cos(t)dt resembles the formula:int x^3 cos(ax) dx = ((3x^2)/a^2-6/a^4)cos(ax)+(x^3/a-(6x)/a^3)sin(ax) .
By comparison, the corresponding values are: x=t and a=1 . Applying the corresponding values on the formula, we get:
int_0^(pi/2) t^3 cos(t)dt=[((3t^2)/1^2-6/1^4)cos(1*t)+(t^3/1-(6t)/1^3)sin(1*t)]|_0^(pi/2)
=[(3t^2-6)cos(t)+(t^3-6t)sin(t)]|_0^(pi/2)
To solve for the definite integral, we may apply the formula: F(x)|_a^b = F(b)-F(a) .
[(3t^2-6)cos(t)+(t^3-6t)sin(t)]|_0^(pi/2)
=[(3(pi/2)^2-6)cos(pi/2 )+((pi/2)^3-6(pi/2))sin(pi/2 )]
-[(3(0)^2-6)cos(0)+((0)^3-6(0))sin(0)]

=[(3pi^2/4-6)*0+((pi^3/8-3pi))*1] -[(0-6)*1 +(0-0)*0 ]

=[0+pi^3/8-3pi] -[-6 +0 ]

=[pi^3/8-3pi] -[-6 ]

=pi^3/8-3pi+6 or 0.451 (approximated value)