`int_0^(pi/2) t^3cost dt` Use integration tables to evaluate the definite integral.

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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)

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