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16 year old college math student beginner at calculus.

Radians not degrees for the trigonometric functions.

This area is exactly the same as the area enclosed by each infinitely repeating pattern in the graph $\cos (x) + \cos (y) = 1$

Graph of $\sin (x) + \sin (y) = 1$:

My working:

  • This area is exactly the same as the area enclosed by each infinitely repeating pattern in the graph $\cos (x) + \cos (y) = 1$
  • For cosine, the pattern is split symmetrically across all four quadrants of the graph around the origin so integration may be possible to find half of the desired area
  • Solve for y:
    • Subtracting $\cos (x)$ from both sides of the equation: $\cos (y) = 1 - \cos (x)$
    • Taking inverse cosine of both sides of the equation: $y = \arccos (1 - \cos (x))$
  • Since the range of the inverse cosine function is $-1 ≤ \arccos (x) ≤ 1$, if the value of $\cos (x) < 0$, the value in within the parentheses will be outside of the inverse cosine range and not be valid, so the new function has a minimum of $\arccos(1) = 0$, likewise, if the maximum y value of $\cos (x)$ would be 1, therefore the other extreme point, the maximum, of the new function would be $\arccos (0) = \frac{\pi}{2}\cdot$ Because any resulting definite integral of this function between any two valid points on the x-axis will be positive, we don't have to consider translating the graph in order to find a more appropriate area-under-the-curve value.
  • We know that, from the graph, provided, that the curve of the function between two valid points will be an arch-shape, similar to the original wave function graphs. So we know that the desired area will be located between two 'paired' roots of the new function, which we can use as parameters for the indefinite integral later.
  • Find two subsequent roots for the function:
    • Set $y = 0$, so the equation to solve is $\arccos (1 - \cos (x)) = 0$
    • Take cosine of both sides of the equation: $1 - \cos (x) = \cos (0) = 1$
    • Replace the subtrahend with the value on the right hand side of the equation: $\cos (x) = 1 - 1 = 0$
    • Take inverse cosine of both sides: $x = \arccos (0) = \frac{π}{2}$
    • Since cosine is an even function, all positive roots can be turned negative and still be valid: $\cos (-\frac{π}{2}) = 0$ confirms this in this context.
  • Since the y-intercept is a maximum y-value of the function, we can assume that the shape of the function's curve on both sides of the y-axis are symmetrical and that we have the correct parameters, the diameter of the shape in the graph also seems reasonable and correlates with this conclusion.
  • Integrate this definite integral: $\displaystyle\int_{-\frac{π}{2}}^{\frac{π}{2}} \! \arccos(1 - \cos (x)) \, \mathrm{d}x.$
    • Take the antiderivative of the function first:
      • $\int \arccos (1 - \cos (x)) dx$
      • Use u-substitution (only way I know how to integrate) for the contents inside the parentheses: $1 - \cos (x) = u$
      • Solve for x:
        • Replace the subtrahend with the value on the right hand side of the equation: $\cos (x) = 1 - u$
        • Take inverse cosine of both sides: $x = \arccos (1 - u)$
      • Find dx:
        • Differentiate both sides of the equation: $du = \frac{d}{dx} (1 - \cos (x)) = \frac{d}{dx} (1) + \frac{d}{dx} (-\cos(x)) = 0 + \sin (x) = \sin(x)$ --> $du = \sin (x)dx$
        • Divide both sides by $\sin (x)$ to find $dx$: $dx = \frac{du}{\sin (x)}$
      • Substitute the u value into the remaining x from the $dx$ expression: $dx = \frac{du}{\sin (\arccos (1 - u))}$
      • Simplify the new u-substituted antiderivative expression: $\frac{\int \arccos (u) du}{\sin (\arccos (1-u))}$
      • Integrate the antiderivative in the numerator: $\frac{u\arccos (u) + \sqrt{1 - u^2}}{\sin (\arccos (1-u))}$
      • Resubstitute the x-values back into this new expression: $\frac{(1 - \cos (x))(\arccos (1 - \cos (x)) + \sqrt{1 - (1 - \cos (x))^2}}{\sin (\arccos (\cos (x))}\cdot$
      • Simplify this expression: \begin{eqnarray*} &&\kern-2em\frac{\arccos (1 - \cos (x)) - \cos (x)\arccos (1 - \cos (x)) + \sqrt{1 - (1 - 2\cos (x) + \cos^{2}(x))}}{\sin ( \arccos (\cos (x))}\\ &=&\frac{\arccos (1 - \cos (x)) - \cos (x)\arccos (1 - \cos (x)) + \sqrt{2\cos (x) - \cos^{2}(x)}}{\sin ( \arccos (\cos (x))}\cdot \end{eqnarray*}
    • Use the definite integral formula $\displaystyle\int_a^b f(x)\;\mathrm{d}x = F(b) - F(a)$: \begin{eqnarray*} &&\frac{\arccos (1 - \cos (\frac{\pi}{2})) - \cos (\frac{\pi}{2})\arccos (1 - \cos (\frac{\pi}{2})) + \sqrt{2\cos (\frac{\pi}{2}) - \cos^{2}(\frac{\pi}{2})}}{\sin ( \arccos (\cos (\frac{\pi}{2}))}\\ &&- \frac{\arccos (1 - \cos (-\frac{\pi}{2})) - \cos (-\frac{\pi}{2})\arccos (1 - \cos (-\frac{\pi}{2})) + \sqrt{2\cos (-\frac{\pi}{2}) - \cos^{2}(-\frac{\pi}{2})}}{\sin ( \arccos (\cos (-\frac{\pi}{2}))}\cdot \end{eqnarray*}
  • This simplifies to $0 - 0 = 0$, but it's not zero because I can see that the the area inside the pattern is a finite value, therefore half the area (the definite integral at hand) is a finite value. I don't know where I went wrong or what to do.
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    $\begingroup$ It turns out that you performed the $u$-substitution incorrectly: the factor you're dividing by must go inside the integral, not outside. WolframAlpha cannot integrate $\arccos(1-\cos x)$, so maybe there is no elementary antiderivative. $\endgroup$ Jan 22, 2023 at 0:26
  • $\begingroup$ Maple also finds no elementary antiderivative. Numerically, $\int_{-\pi/2}^{\pi/2} \arccos(1-\cos(x))\; dx \approx 3.64744119225206997400916081676$. $\endgroup$ Jan 22, 2023 at 6:06
  • $\begingroup$ Do you only need the definite integral or an approximation of the antiderivative ? $\endgroup$ Jan 22, 2023 at 9:10
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    $\begingroup$ If I just needed an approximation I would use a ruler. $\endgroup$
    – user1112591
    Jan 22, 2023 at 12:26
  • $\begingroup$ I'm pretty sure the definite integral has a closed-form in terms of the generalized hypergeometric function $_4F_3$. If you were hoping for a simpler final result in terms of recognizable functions, you're probably out of luck. $\endgroup$
    – David H
    Jan 22, 2023 at 12:37

3 Answers 3

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Your area uses $\cos^{-1}(y)$’s series which converges on $|y|<1$ with the Pochhammer symbol $(u)_v$:

$$\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=2\int_0^\frac\pi2\cos^{-1}\left(2\sin^2\left(\frac x2\right)\right)dx=2\int_0^\frac\pi2\frac\pi2-\sum_{n=0}^\infty\frac{\left(2\sin^2\left(\frac x2\right)\right)^{2n+1}\left(\frac12\right)_n}{(2n+1)n!}dx$$

Switching operators gives:

$$\frac{\pi^2}2-\sum_{n=0}^\infty\frac{2^{2n+2}\left(\frac12\right)_n}{(2n+1)n!}\int_0^\frac\pi2 \sin^{4n+2}\left(\frac x2\right)dx= \frac{\pi^2}2-\sum_{n=0}^\infty\frac{2^{2n+1}\left(\frac12\right)_n}{(2n+1)n!}\int_0^\frac\pi4 \sin^{4n+2}(x)dx $$

Next, use the $\int \sin^r(x)$ to incomplete beta B$_x(a,b)$ identity.

$$\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=\frac{\pi^2}2-\sum_{n=0}^\infty\frac{\left(\frac12\right)_k4^{n+1}}{(2n+1)n!}\text B_\frac12\left(\frac32+2n,\frac12\right)$$

Finally, use a B$_x(a,b)$ series, switch the series, and simplify with hypergeometric $_4\text F_3$:

$$\boxed{\int_{-\frac\pi2}^\frac\pi2\cos^{-1}(1-\cos(x))dx=\frac1{\sqrt\pi}\left(4\Gamma^2\left(\frac34\right)\,_4\text F_3\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;\frac14\right)+\frac{\Gamma^2\left(\frac14\right)}{36}\,_4\text F_3\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;\frac14\right)\right)= 3.64744119225206997400916081676416139843462\dots}$$

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  • $\begingroup$ If I left my answer as the integral $\displaystyle\int_{-\frac{π}{2}}^{\frac{π}{2}} \! \arccos(1 - \cos (x)) \, \mathrm{d}x.$, would I have to prove that this integral has no elementary solutions? How would I do that? Is leaving it as the integral technically a 'closed form' or not, after all, differentiating and integrating are just forms of simplifying expressions (the things on the two sides of the equal sign of the calculus are the same thing so the calculus is just a simplification of expressions). $\endgroup$
    – user1112591
    Jan 22, 2023 at 14:54
  • $\begingroup$ Thank you for your time. Is there a way in the future that I can immediately check if an antiderivative gives closed form/elementary solutions or not? The equivalent of the quadratic discriminant but with antiderivatives. $\endgroup$
    – user1112591
    Jan 22, 2023 at 15:14
  • $\begingroup$ @AshtonDowling There is a great post here and here $\endgroup$ Jan 22, 2023 at 15:23
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We can make a decent approximation of the antiderivative using one of my favored $\large 1,400$ years old approximations $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$

This gives $$I_1=\int \cos ^{-1}(1-\cos (x))\,dx \sim \int \cos ^{-1}\left(\frac{5 x^2}{x^2+\pi ^2}\right)\,dx=I_2$$ To check, consider the norm $$\Phi=\int_{-\frac \pi 2}^{+\frac \pi 2}\Bigg[\cos ^{-1}(1-\cos (x))-\cos ^{-1}\left(\frac{5 x^2}{x^2+\pi ^2}\right) \Bigg]^2\,dx=6.145\times 10^{-6}$$

$$I_2=x \cos ^{-1}\left(\frac{5 x^2}{x^2+\pi ^2}\right)-5 i \sqrt{\frac{2}{3}} \pi \,\large A $$ where $${ \large A}= F\left(i \sinh ^{-1}\left(\frac{\sqrt{6} }{\pi }x\right)|-\frac{2}{3}\right)-\Pi \left(\frac{1}{6};i \sinh ^{-1}\left(\frac{\sqrt{6} }{\pi }x\right)|-\frac{2}{3}\right)$$ Some results for the integration between $0$ and $t$ $$\left( \begin{array}{ccc} t &\text{approximation} & \text{exact} \\ 0.1 & 0.1569109 & 0.1569131 \\ 0.2 & 0.3128116 & 0.3128286 \\ 0.3 & 0.4667036 & 0.4667585 \\ 0.4 & 0.6176100 & 0.6177321 \\ 0.5 & 0.7645818 & 0.7648015 \\ 0.6 & 0.9066998 & 0.9070419 \\ 0.7 & 1.0430706 & 1.0435487 \\ 0.8 & 1.1728158 & 1.1734269 \\ 0.9 & 1.2950534 & 1.2957741 \\ 1.0 & 1.4088677 & 1.4096524 \\ 1.1 & 1.5132609 & 1.5140428 \\ 1.2 & 1.6070722 & 1.6077677 \\ 1.3 & 1.6888294 & 1.6893476 \\ 1.4 & 1.7563969 & 1.7566577 \\ 1.5 & 1.8059298 & 1.8058995 \\ \end{array} \right)$$

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$\int_{-\frac{π}{2}}^{\frac{π}{2}} \! \arcsin(1 - \sin(x)) \, \mathrm{d}x = \int_{-\frac{π}{2}}^{0} \! \arcsin(1 - \sin (x)) \, \mathrm{d}x + \int_{0}^{\frac{π}{2}} \! \arcsin(1 - \sin(x)) \, \mathrm{d}x$

Now, using the substitution $x\to\frac{π}{2} -x$ on the two terms in $RHS$ of above equation, we have,

$\int_{-\frac{π}{2}}^{\frac{π}{2}} \! \arcsin(1 - \sin (x)) \, \mathrm{d}x = \int_{-\frac{π}{2}}^{0} \! \arcsin(1 + \cos (x)) \, \mathrm{d}x + \int_{0}^{\frac{π}{2}} \! \arcsin(1 - \cos (x)) \, \mathrm{d}x$

Now, you say that

This area is exactly the same as the area enclosed by each infinitely repeating pattern in the graph $\cos(x)+\cos(y)=1$

Assuming this as true and the fact that $\arcsin(u) + \arccos(u) = π/2$ for $u$ be any thing in suitable domain , can you do rest by yourself ?

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  • $\begingroup$ No, I can't, that splitting the definite integral parameters thing you did just makes the problem even more complicated, and the definite integral isn't the problem here, it's that I can't find the antiderivative of the function. I don't know how to do integration by parts, and by the looks of it I don't know how to do u-substitutions either. $\endgroup$
    – user1112591
    Jan 22, 2023 at 12:37
  • $\begingroup$ @AshtonDowling Do you know math.stackexchange.com/q/3856735/1020656 ? $\endgroup$ Jan 22, 2023 at 12:38
  • $\begingroup$ No need to evaluate indefinite integral. $\endgroup$ Jan 22, 2023 at 12:40
  • $\begingroup$ @An_Elephant Did not downvote. Why does your integral maybe give a complex number? Perhaps this $\cos^{-1}(t)$ integral $\endgroup$ Jan 22, 2023 at 13:44
  • $\begingroup$ @TymaGaidash Sorry, I will try to look it further to find the flaw and edit it immediately. $\endgroup$ Jan 22, 2023 at 14:21

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