Help with inverse integration (1 Viewer)

z3bra · May 24, 2017

Question 23 please

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WildestDreams · May 24, 2017

Wouldn't it be sin-1x - the graph of the tangent?

But to find the equation of the graph you'll need to find the tangent of sin-1x when x= √3/2

InteGrand · May 24, 2017

WildestDreams said:
Wouldn't it be sin-1x - the graph of the tangent?

$\noindent Yes, but they don't know how to integrate $\sin^{-1} x$ in 3U. So they would be expected to use other methods.$

WildestDreams · May 24, 2017

InteGrand said:
$\noindent Yes, but they don't know how to integrate $\sin^{-1} x$ in 3U. So they would be expected to use other methods.$

So what method would that be?
tangent line - sin-1x
and then change the inverse of sin-1x to a normal function, but calculating it terms of that integral?

pikachu975 · May 24, 2017

From first look you can't integrate sin^-1 x so find the area of the rectangle and minus two integrals which are the white spaces.

For the top white space integrate in terms of y so x = siny and now this can be integrated. The bottom one should be easy just a triangle so an integration isn't needed.

Seems like first you need to find gradient and equation of the tangent then find its x intercept.

z3bra · May 24, 2017

Okay. I'll try.

z3bra · May 24, 2017

Can u please show working out.

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pikachu975 · May 24, 2017

z3bra said:
Can u please show working out.

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I forgot that there was an area under the x axis so never mind.

kawaiipotato · May 25, 2017

z3bra said:
Question 23 please

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It's highly recommended to draw a diagram when using integration to find an area (and if it's given, try drawing it yourself as an exercise (unless you're time constricted)!).

$\noindent Notice that for a defined inverse function $f^-1$ of $f$, the domain and ranges are swapped around. So the area A bounded by $y = sin^{-1}x$ and the tangent to $y$ at $(\frac{\sqrt{3}}{2},\frac{\pi}{3})$ corresponds to the area bounded by $y = sinx$ and the tangent at $(\frac{\pi}{3},\frac{\sqrt{3}}{2})$ (try drawing a diagram of $y = sin^{-1}x$ and $y=sinx$ to make it clearer. \\From the diagram with $y=\sin x$, the tangent to the curve $y=\sin x$ at $(\frac{\pi}{3},\frac{\sqrt{3}}{2})$ is $y = \frac{\sqrt{3}}{2} + \frac{1}{2} ( x - \frac{\pi}{3})$. \\ Then the area is just $A = \int_{\frac{\pi}{3} - \sqrt{3}}^{\frac{\pi}{3}} \frac{\sqrt{3}}{2} + \frac{1}{2} ( x - \frac{\pi}{3}) dx - \int_{0}^{\frac{\pi}{3}} sinx dx$ which evaluates to $A = \frac{1}{4}$. (The first integral is just the area under the tangent, while the second is the area under $sinx$, which gives the remaining area that was to be found).$
(Integration limits were found by inspection of the diagram of y=sinx and its tangent).

Help with inverse integration (1 Viewer)

z3bra

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WildestDreams

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InteGrand

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WildestDreams

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pikachu975

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z3bra

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z3bra

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pikachu975

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kawaiipotato

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