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

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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**

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