Locating an enhancement formula without trigonometry

I'm attempting to recognize far better the adhering to enhancement formula: $$\int_0^a \frac{\mathrm{d}x}{\sqrt{1-x^2}} + \int_0^b \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \int_0^{a\sqrt{1-b^2}+b\sqrt{1-a^2}} \frac{\mathrm{d}x}{\sqrt{1-x^2}}$$

The term $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ can be stemmed from trigonometry (given that $\sin(t) = \sqrt{1 - \cos^2(t)}$) yet I have not had the ability to locate any kind of means to acquire this formula without trigonometry, just how could it be done?

modify: dealt with a blunder in my formula.

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2019-05-06 23:20:33
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Answers: 2

Replace the first indispensable by the very same point from $-a$ to $0$, and also take into consideration the factors W, X, Y, Z on the device circle over $-a,0,b$ and also $c = a\sqrt{1-b^2} + b \sqrt{1-a^2}$. Attract the family members of lines alongside XY (and also WZ). This family members establishes a map from the circle to itself ; via each factor, attract a parallel and also take the various other junction of that line with the circle.

Your formula claims that this map [edit : or instead the map it generates on the $x$ - works with of factors on the circle ] is an adjustment of variables transforming the indispensable on $[-a,0]$ to the very same indispensable on $[b,c]$. Whatever distinction you execute in the procedure of confirming this, will certainly be the confirmation that $dx/y$ is a turning - stable differential on the circle $x^2 + y^2 = 1$.

[The generated map on x - works with is : $x \to$ factor on semicircle over $x \to$ equivalent factor on line alongside XY $\to x$ - coordinate of the 2nd factor. Below were are simply recognizing $[-1,1]$ with the semicircle over it. ]

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2019-05-08 20:37:48
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I will certainly attempt to sum up below what I've located until now on this :

Since the kind being incorporated is $\frac{\mathrm{d}x}{\sqrt{1 - x^2}} =: \omega$ allow $y = \sqrt{1 - x^2}$, we can additionally write this as $x^2 + y^2 - 1 = 0$ that makes it extra noticeable that this is a device circle.

Allow $N$ be the factor $(1,0)$ and also we can specify a team framework on the contour as adheres to. To add $A$ and also $B$, fire a ray from $N$ alongside $AB$ and also select the factor it converges the contour as $A \oplus B$. Symbolically, this suggests $A \oplus B = N - k(B - A)$ for some nonzero $k$, given that $A \oplus B$ pushes the circle we can expand it right into the formula of the contour to address for $k$,

$$\begin{align} (- k (x_b - x_a))^2 + (1 - k (y_b - y_a))^2 - 1 &= 0 \\\\ k ((x_b - x_a)^2 + (y_b - y_a)^2) - 2 (y_b - y_a) &= 0 \\\\ \frac{2 (y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} &= k \end{align}$$

therefore $x_{a \oplus b} = - 2 \frac{(x_b - x_a)(y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} = x_a \sqrt{1-x_b^2} + x_b \sqrt{1-x_a^2}$.

Currently I assume by the theorem, if

  • $\omega$ is stable along the contour
  • $\oplus$ is the team regulation on factors of the contour

we can end since

$$\int_0^a \omega + \int_0^b \omega = \int_0^{a \oplus b} \omega$$

Certainly, the previous formula holds - but also for this blog post to count as an evidence the theory is required. In addition I think this very same suggestion benefits any kind of conic area as well as additionally for cubic contours and also probably none others contours.

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2019-05-08 05:15:31
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