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I am trying to find the area of the shaded area

I am trying to find the area of the shaded area:

I formed a rectangle with width of $10\sqrt{2}$ and length of $\frac{\pi}{2}$

Then I need to subtract from the area under graph of $y=5 \csc x \cot x$

To tackle this, I solve part by part $\int_{\pi/4}^{\pi/2} y$ + ?

$\int_{\pi/4}^{\pi/2} y$ is bounded by the x axis so its easy to find the area under the graph. but what about from $\frac{\pi}{2}$ to $\frac{3 \pi}{4}$ ? its bounded by nothing and the area goes to infinity. How do I find the area ?

Blue
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user307640
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    The curve for $ \ 5 \csc x \cot x \ $ lies below the horizontal line $ \ y \ = \ 5 \sqrt2 \ \ , \ $ so you want to integrate over the interval $ \ \left(\pi/4 \ , \ 3\pi/4 \right) \ $ with $ \ 5 \sqrt2 \ $ being the "upper curve" and $ \ 5 \csc x \cot x \ $ being the "lower curve". The integration is simple since the integrand is the derivative of a trig function you can look up. (If you observe the symmetry of the curve, when drawn to scale, you don't really need to integrate at all!) –  Oct 18 '22 at 03:31
  • @boojum what do you mean by "upper curve" and "lower curve", I do not know how to put it into a mathematical expression – user307640 Oct 18 '22 at 03:37
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    The "area between two curves" is $ \ \int_a^b \ f_{upper}(x) \ - \ f_{lower}(x) \ \ dx \ \ . $ –  Oct 18 '22 at 03:38
  • @boojum For learning and understanding purposes, is this linked to any properties of definite integrals? or etc – user307640 Oct 18 '22 at 03:40
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    Over the interval spanned by your "blue region", the function $ \ 5 \csc x \cot x \ $ is less than or equal to $ \ 5 \sqrt2 \ \ . \ $ So you want to integrate the differences in $ \ y \ $ between the horizontal line upper boundary and the curved lower boundary. "Area between two curves" is a topic that should be covered under definite integration. –  Oct 18 '22 at 03:43

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The crisp way to solve this is to note that $5\csc\theta\cot\theta$ is an odd function about $\pi/2$: $$5\csc(\pi-\theta)\cot(\pi-\theta)=-5\csc\theta\cot\theta$$ So the parts above and below the $x$-axis can be combined into a rectangle with width $\pi/2$ and height $5\sqrt2$, hence area $5\pi/\sqrt2$.

The mechanical way is to notice that the antiderivative of $\csc x\cot x=\cos x/\sin^2x$ is $-\csc x$, so the area is $$\int_{\pi/4}^{3\pi/4}(5\sqrt2-5\csc\theta\cot\theta)\,d\theta=5\pi/\sqrt2-5[-\csc\theta]_{\pi/4}^{3\pi/4}=5\pi/\sqrt2-5(-1/\sqrt2-(-1/\sqrt2))=5\pi/\sqrt2$$

Parcly Taxel
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From $\frac\pi4$ to $\frac\pi2$, area is bounded by $x$ axis, so, you write $\int_\frac\pi4^\frac\pi2ydx$.

Similarly, from $\frac{3\pi}4$ to $\frac\pi2$ (note the reversed limits), area is bounded by $x$ axis, so, we write $\int_{\frac{3\pi}4}^\frac\pi2ydx$.

aarbee
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