I have difficulty on how to eliminate parameter especially the equation involved trigonometry equation.

The question is asking for the area bounded by the curve , the 2 axes and the line $y=1$.

$x=4 \sin t$

$y=\cot t$

where $t$ is in the range $(0, \pi)$. I have tried to use the trigonometric formula $1+ \cot^2 t =\csc^2 t$

but I got the wrong answer.The answer given is 4ln[(square root 2)+1].

  • $\begingroup$ can you draw the curve? $\endgroup$ – jimjim Mar 1 '19 at 5:47
  • $\begingroup$ The curve is not given by the question. $\endgroup$ – Lee Jason Mar 1 '19 at 6:01
  • $\begingroup$ yes it is, it is parametrically given, can you draw it or not? $\endgroup$ – jimjim Mar 1 '19 at 6:03
  • $\begingroup$ no, I don't know how to draw $\endgroup$ – Lee Jason Mar 1 '19 at 6:06
  • $\begingroup$ put t=0 and find x and y, put t = 0.1 and find x and y, do a few steps or use wolfram alpha to draw it for you, it is fun. $\endgroup$ – jimjim Mar 1 '19 at 6:15

The Desmos plot of the relevant part of the curve looks like this:

The desired area is simply the area "under" (to the left of) the curve from $y=0$ to $y=1$.

We now express $x$ in terms of $y$. Since $1+\cot^2t=\csc^2t=\frac1{\sin^2t}$ and thus $\frac1{1+\cot^2t}=\sin^2t$, we have $$x=4\sin t=\frac4{\sqrt{1+\cot^2t}}=\frac4{\sqrt{1+y^2}}$$ So the desired area is $$\int_0^1\frac4{\sqrt{1+y^2}}\,dy=4[\sinh^{-1}y]_0^1=4\sinh^{-1}1=4\ln(1+\sqrt2)$$ The last equality is well-known; see e.g. OEIS A091648.


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