Calculus with Vector Functions — Question 5

Question 5 – Solution

Strategy. Integrate componentwise. The first two components require integration by parts; the third can be checked in two different ways.

Step 1: First component Integration by parts with $u=t$, $dv=\cos t\,dt$ gives $$\begin{align*} \int_0^{\pi/2}t\cos t\,dt &=\left[t\sin t\right]_0^{\pi/2}-\int_0^{\pi/2}\sin t\,dt\\ &=\frac\pi 2-1. \end{align*}$$

Step 2: Second component With $u=t$, $dv=\sin t\,dt$, $$\begin{align*} \int_0^{\pi/2}t\sin t\,dt &=\left[-t\cos t\right]_0^{\pi/2}+\int_0^{\pi/2}\cos t\,dt=1. \end{align*}$$

Step 3: Third component Since $d(\sin^2t)/dt=2\sin t\cos t$, $$\int_0^{\pi/2}\sin t\cos t\,dt =\left[\frac 12\sin^2t\right]_0^{\pi/2}=\frac 12.$$ Therefore $$\boxed{\int_0^{\pi/2}\left\langle t\cos t,t\sin t,\sin t\cos t\right\rangle\,dt =\left\langle \frac\pi 2-1,1,\frac 12\right\rangle}.$$ Using $\sin t\cos t=\tfrac 12\sin 2t$ gives the same third component: $[-\tfrac 14\cos 2t]_0^{\pi/2}=1/2$.