What is the distance if the velocity is v=1/sin^2t*cos^2t

The distance is ds =
v*dt.

We'll integrate both
sides:

Integral ds = Integral
v*dt

s = Integral v*dt

To
integrate the given function, we'll have to re-write the numerator of the function to
give odds to the resolution of the indefinite
integral.

We'll substitute the numerator1, by the
fundamental formula of trigonometry:

(sin t)^2 + (cos
t)^2 = 1

We'll re-write the
ratio:

1/(sin t)^2*(cos t)^2=[(sin t)^2 + (cos t)^2]/(sin
t)^2*(cos t)^2

1/(sin t)^2*(cos t)^2 = (sin t)^2/(sin
t)^2*(cos t)^2 + (cos t)^2/(sin t)^2*(cos t)^2

We'll
simplify the ratios:

1/(sin t)^2*(cos t)^2 = 1/(cos t)^2 +
1/(sin t)^2

We'll integrate both
sides:

Int dt/(sin t)^2*(cos t)^2 = Int dt/(cos t)^2 + Int
dt/(sin t)^2

Int dt/(sin t)^2*(cos t)^2 =
tan t - cot t + C

The
expression of the function of distance
is:

s(t) = tan t -
cot t