# Indefinite integral with sector of ellipse

An ellipse is given by the following equation: $$152 x^2 - 300 x y + 150 y^2 - 42 x + 40 y + 3 = 0$$ After solving for the midpoint we have: $$152 (x-1/2)^2 - 300 (x-1/2) (y-11/30) + 150 (y-11/30)^2 = 1/6$$ Introducing polar coordinates: $$x = 1/2 + r \cos(\theta) \quad ; \quad y = 11/30 + r \sin(\theta)$$ Giving: $$152\, r^2 \cos^2(\theta) - 300\, r^2\, \cos(\theta) \sin(\theta) + 150\, r^2 \sin^2(\theta) = 1/6 \quad \Longrightarrow \\ \frac{1}{2} r^2(\theta) = \frac{1/12} {152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}$$ The area of a sector of the ellipse is: $$\int_{\theta_1}^{\theta_2} \frac{1}{2} r^2(\theta) \, d\theta$$ So it seems that we have to find the indefinite integral: $$\int \frac{1/12 \, d\theta} {152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}$$ And then I'm stuck. Because feeding this into MAPLE with

int(1/12/(152*cos(theta)^2-300*cos(theta)*sin(theta)+150*sin(theta)^2),theta);

quite to my surprise, gives a complex result: $${\frac {1}{720}}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta \right) \right) ^{2}+ \left( {\frac {5}{38}}\,i\sqrt {3}+{\frac {75} {38}} \right) \tan \left( 1/2\,\theta \right) -1 \right) \\ -{\frac {1}{720 }}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta \right) \right) ^{2}+ \left( -{\frac {5}{38}}\,i\sqrt {3}+{\frac {75}{38}} \right) \tan \left( 1/2\,\theta \right) -1 \right)$$ But I'm pretty sure that the area of an ellipse sector is a real number. So the question is: does there exist a closed form for the abovementioned integral that is real valued instead of complex? What is it? And why that complex result with MAPLE?

$$A = (1/2,3/10) \quad ; \quad B = (1/2,1/3) \quad ; \quad C = (1/2,11/30) \\ P = (1/3,1/6) \quad ; \quad Q = (2/3,1/2) \\ R = (1/4,1/10) \quad ; \quad S = (3/4,3/5)$$ Triangle edges are black; ellipse is $\color{red}{red}$ .
Triangle areas: $\Delta PAQ = 1/180$ , $\Delta PRB = \Delta QSB = 1/720$ .
It is conjectured that the ellipse sector area $\overline{CRBSC}$ is exactly $1/3$ of the total ellipse area;
the latter being $= \pi\sqrt{3}/180$ . Can someone prove or disprove this conjecture?

• You have most of the necessary information for finding a linear (well, affine) mapping that takes the ellipse to, for example, the standard unit circle; your rays at angles $\theta_1$ and $\theta_2$ are mapped to two radii, area between explicit. Now map back, area is constant multiple. – Will Jagy Dec 1 '14 at 23:24
• If you don't like that, any rational function of sine and cosine c an be transformed into a rational function using the Weierstrass substitution, en.wikipedia.org/wiki/Tangent_half-angle_substitution – Will Jagy Dec 1 '14 at 23:26

$$\int\frac{dx}{a\cos^2x-b\cos x\sin x+c\sin^2x}=\frac2{\sqrt\Delta}\cdot\tanh^{-1}\bigg(\frac{b-2c\tan x}{\sqrt\Delta}\bigg)$$ where $$\Delta=b^2-4ac$$. If $$\Delta<0$$, just use Euler's formula to transform the hyperbolic arctangent of complex argument into a trigonometric one of real argument. For $$\Delta=0\iff b=\pm2\sqrt{ac}$$,
we get $$I=\dfrac{b+(a+c)\sin2x}{(a+c)\Big[(a-c)+(a+c)\cos2x\Big]}$$.
• In our case $b^2-4ac < 0$ because we have an ellipse. Does that mean that the outcome is:$$-\frac2{\sqrt\Delta}\cdot\tan^{-1}\bigg(\frac{b-2c\tan x}{\sqrt\Delta}\bigg)$$ with $\Delta = 4ac-b^2 > 0$ ? – Han de Bruijn Dec 2 '14 at 11:24
• @HandeBruijn: Yes, exactly. $($ Mathematica agrees also $)$. – Lucian Dec 2 '14 at 12:10
• Thanks, Lucian: I can calculate now the area of the desired sector. With the surprising result that it seems to be exactly $1/3$ of the total area of the ellipse ! Can't figure out why .. – Han de Bruijn Dec 2 '14 at 17:36
The integral is elementary. $$I = \int \frac{d \theta} {a \cos^2(\theta) - b \cos(\theta) \sin(\theta) + c \sin^2(\theta)}= \int \frac{d \tan(\theta)} {a - b \tan(\theta) + c \tan^2(\theta)}$$ Let $u=\tan(\theta)$ , then: $$I = \int \frac{d u}{c u^2 - b u + a} = \int \frac{d u/c}{\left[ u-b/(2c) \right]^2+a/c - \left[b/(2c)\right]^2}=\\ 2 \int \frac{d (2c u-b)}{(2c u-b)^2+(4ac-b^2)}= \frac{2}{\sqrt{4ac-b^2}} \int \frac{d\left[(2c u-b)/ \sqrt{4ac-b^2}\right]} {1+\left[(2c u-b)/\sqrt{4ac-b^2}\right]^2}$$ Now let $\Delta = \sqrt{4ac-b^2}$ and we're finished: $$I(\;\tan(\theta)\;) = \frac{2}{\Delta} \arctan\left(\frac{2c\tan(\theta)-b}{\Delta}\right)$$ Fill in the numbers for our sector: $a = 152\;,\;b = 300\;,\;c = 150$ , $\tan(\theta_2) = 16/15\;,\;\tan(\theta_1) = 14/15$ ; remember that an $\arctan$ is only defined for arguments $\in \left[-\pi/2,+\pi/2\right]$ and that the whole area of the ellipse equals $\pi\sqrt{3}/180$ . Then: $$\mbox{area} = \frac{\pi\sqrt{3}}{180}/2 - \left[\;I(16/15)-I(14/15)\;\right]/12 = \frac{\pi\sqrt{3}}{540}$$ Thus establishing that the AfterMath conjecture is true. So finally we have all the ingredients to calculate the area of the "ellipse with a hat" ( $\color{blue}{blue} + \color{red}{red} + \color{green}{green}$ ) : $$\color{blue}{\pi\sqrt{3}/180(1-1/3)+1/120}+\color{red}{2/720}+\color{green}{1/180} = \frac{\pi\sqrt{3}}{270}+\frac{1}{60}$$