Tan Particle Motion

$\begin{aligned} a & = \tan t \\ \implies \color{#3D99F6} v & = \int \tan t \ dt & \small \color{#3D99F6} \text{where } v = \frac {dx}{dt} \text{ is the velocity.} \\ & = \int \frac {\sin t}{\cos t} dt \\ & = - \int \frac {d \cos t}{\cos t} \\ & = - \ln (\cos t) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \because v(0) & = - \ln (\cos 0) + C = 0 &\small \color{#3D99F6} \implies C = 0 \\ \implies v & = - \ln (\cos t) \end{aligned}$

Therefore, the distance traveled after $\frac \pi 2$ s:

$\begin{aligned} x \left(\frac \pi 2\right) & = - \int_0^\frac \pi 2 \ln (\cos t) \ dt & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - \frac 12 \int_0^\frac \pi 2 (\ln (\cos t) + \ln (\sin t)) \ dt \\ & = - \frac 12 \int_0^\frac \pi 2 \ln (\sin t\cos t) \ dt \\ & = - \frac 12 \int_0^\frac \pi 2 \ln \left(\frac 12 \sin (2t)\right) \ dt \\ & = - \frac 12 \int_0^\frac \pi 2 (\ln (\sin ({\color{#3D99F6}2t})) - \ln 2 ) \ dt & \small \color{#3D99F6} \text{Let }u = 2t \implies du = 2\ dt \\ & = - {\color{#3D99F6}\frac 14 \int_0^\pi \ln (\sin u)\ du} + \frac 12 \int_0^\frac \pi 2 \ln 2 \ dt & \small \color{#3D99F6} \text{Since }\sin u \text{ is symmetrical about }\frac \pi 2 \\ & = - {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\sin u)\ du} + \frac {t \ln 2}2 \ \bigg|_0^\frac \pi 2 & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\cos u)\ du} + \frac {\pi \ln 2}4 \\ & = \frac 12 x \left(\frac \pi 2\right) + \frac {\pi \ln 2}4 \\ & = \frac {\pi \ln 2}2 \end{aligned}$

Therefore, $ab = \frac 12 \times 2 = \boxed{1}$ .

The velocity is $v = \ln(\sec t)$ , and so the distance travelled is $\int_0^{\frac12\pi} \ln(\sec t)\,dt \; = \; -J$ where $J \; = \; \int_0^{\frac12\pi} \cos t\,dt \; = \; \int_0^{\frac12\pi} \sin t \,dt$ and hence $\begin{aligned} 2J & = \; \int_0^{\frac12\pi} \sin t \cos t\,dt \; = \; \int_0^{\frac12\pi} \tfrac12\sin 2t\,dt \; = \; \int_0^{\frac12\pi} \sin2t\,dt - \tfrac12\pi\ln2 \\ & = \; \tfrac12 \int_0^{\pi}\sin t\,dt - \tfrac12\pi\ln2 \; = \; \tfrac12 \times 2 \int_0^{\frac12\pi} \sin t\,dt - \tfrac12\pi \ln2 \; = \; J - \tfrac12\pi\ln2 \end{aligned}$ so that $J = -\tfrac12\pi\ln2$ , making the distance travelled $\tfrac12\pi\ln2$ , and so the answer is $\tfrac12 \times 2 = \boxed{1}$ .