Why are these primitives containing $\arcsin x$ equal up to a constant?

Question

While trying to solve $\displaystyle\int\sqrt{14x-x^2}\;dx$, I obtained three different primitives in three different ways:

Method 1: completing the square

$c(x)=\dfrac{1}{2}\left[49\arcsin\left(\dfrac{x}{7}-1\right)+(x-7)\sqrt{14x-x^2}\right]$

Method 2: using the substitution $x=u^2$

$s(x)=49\arcsin\left(\sqrt{\dfrac{x}{14}}\right)+\dfrac{1}{2}(x-7)\sqrt{14x-x^2}$

Method 3: Wolfram

$w(x)=\dfrac{1}{2}(x-7)\sqrt{14x-x^2}-49\arcsin\left(\sqrt{1-\dfrac{x}{14}}\right)$

At first, I thought some mistakes were made, as I was not able to reconcile the different appearances of the $\arcsin(\cdot)$ terms. After some inspection using Desmos, it turns out all 3 functions are equal up to a constant. In fact, they are separated by an amount of $\dfrac{49\pi}{4}$:

All three functions (without $+C$):

All three functions (with the appropriate $+C$):

So at this point, I'm pretty much convinced all three methods agree and are valid. However, I would like to know how exactly one can prove the $49\pi/4$ spacing (or $49\pi/2$ depending on which functions are compared) between each primitive, other than using this graphical method, which is much weaker after all.

Bonus 1: why is it that each method leads to a different primitive a priori?

Bonus 2: what method was used to obtain $w(x)$ (Wolfram's primitive) if not the other two?

Thanks in advance for all your input!

Answer 1

Omitting the common part $\frac12 (x-7)\sqrt{14x-x^2}$. From the original integral, $0\le x \le 14$.

Let $c_1 = c(x) - \frac12 (x-7)\sqrt{14x-x^2}$, where $-\frac{49\pi}{4} \le c_1 \le \frac{49\pi}{4}$.

$$\begin{align*} c_1 &= \frac12 \cdot 49\arcsin\left(\frac x7-1\right)\\ \sin\left(2\cdot\frac{c_1}{49}\right) &= \frac x7-1\\ \end{align*}$$

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Let $s_1 = s(x) - \frac12 (x-7)\sqrt{14x-x^2}$, where $0 \le s_1 \le \frac{49\pi}{2}$.

$$\begin{align*} s_1 &= 49\arcsin\sqrt{\frac x{14}}\\ \sin^2\left(\frac{s_1}{49}\right) &= \frac x{14}\\ \cos\left(2\cdot\frac{s_1}{49}\right) &= 1-2\sin^2\left(\frac{s_1}{49}\right) = 1-\frac x7\\ \sin\left(2\cdot\frac{s_1 - 49\pi/4}{49}\right) &= \sin\left(2\cdot\frac{s_1}{49}-\frac\pi2\right)\\ &= -\cos\left(2\cdot\frac{s_1}{49}\right)\\ &= \frac x7-1 \end{align*}$$

So taking $c_1 = s_1 - \frac{49\pi}4$ would match the behaviours of $c$ and $s$.

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Let $w_1 = w(x) - \frac12 (x-7)\sqrt{14x-x^2}$, where $-\frac{49\pi}{2} \le w_1 \le 0$.

$$\begin{align*} w_1 &= -49\arcsin\sqrt{1-\frac x{14}}\\ \sin^2\left(-\frac{w_1}{49}\right) &= 1-\frac x{14}\\ \cos\left(-2\cdot\frac{w_1}{49}\right) &= 1-2\sin^2\left(-\frac{w_1}{49}\right) = \frac x7 -1\\ \sin\left(2\cdot\frac{w_1 + 49\pi/4}{49}\right) &= \sin\left(2\cdot\frac{w_1}{49}+\frac\pi2\right)\\ &= \cos\left(2\cdot\frac{w_1}{49}\right)\\ &= \frac x7-1 \end{align*}$$

So taking $c_1 = w_1 + \frac{49\pi}4$ would match the behaviours of $c$ and $w$.

Answer 2

Short answer: The relation between $\sin t$ and $\cos t$ leads to a relation between $\arcsin $ and $\arccos$ of corresponding arguments. Then the formula connecting $\cos (2t)$ and either $\cos t$ or $\sin t$ leads to a relation that lets us rewrite $\arccos$ for an argument in terms of $\arccos$ for half of argument (or double of it, seen reciprocally).

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Details: Let us denote by $X\in[0,1]$ the value $x/14$ (for $x\in[0,14]$), then by $t$ the value of the arcsine computed in $\sqrt{X}=\sqrt{x/14}$. We want to show in the world of functions taken modulo constants: $$ \frac 12\arcsin\left(\frac x7-1\right) \equiv -\arcsin\sqrt{1-\frac x{14}} \equiv \underbrace{\arcsin\sqrt{\frac x{14}}}_{=t} \qquad \text{ modulo constants ,} $$ or in other words $$ \frac 12\arcsin\left(2X-1\right) \equiv -\arcsin\sqrt{1-X} \equiv \underbrace{\arcsin\sqrt X}_{=t} \qquad \text{ modulo constants ,} $$ Then $\sin t=\sqrt X$, $\sin^2 t=X$, $\cos^2 t=1-X$, $\cos t=\sqrt {1-X}$, $t=\arccos\sqrt{1-X}$. The last equivalence is now traced back to the case that the functions $\arccos$ and $-\arcsin$ differ by a constant.

So we want in fact: $$ \frac \pi 2-\frac 12\arccos\left(2X-1\right) = \arccos\sqrt{1-X} = \underbrace{\arcsin\sqrt X}_{=t} \ , $$ and we get also get rid of the (module of) constants. Note that when $X$ varies in $[0,1]$, $\sqrt X$ and $\sqrt {1-X}$ take values in the same intervals, $\arcsin$ and $\arccos$ of such values are in the interval between $0$ and $\pi/2$. Also the left most term takes values in this interval. For $X=0$ we obtain the common value zero. For $X=1$ we obtain the common value $\frac \pi 2$. Since the cosine is bijective seen as a function on $[0,\pi]$ to $[-1,1]$ the equality between the left most and the other terms, both equal to $t$, is equivalent to the the one obtained after applying the cosine on twice these terms. We compute: $$ \begin{aligned} \cos\color{brown} {2\cdot\left(\frac \pi 2-\frac 12\arccos\left(2X-1\right)\right)} &= \cos\left(\pi-\arccos\left(2X-1\right)\right)\\ &=-\cos\arccos\left(2X-1\right)\\ &=-(2X-1)\\ &=1-2X\ ,\\[3mm] \cos \color{brown}{2\cdot t} &=1-2\sin^2 t\\ &=1-2(\sqrt X)^2=1-2X\ . \end{aligned} $$ So the terms in brown are equal.

$\square$

Answer 3

Bonus 2 $$ \begin{aligned} \int \sqrt{14 x-x^2} d x = & \int \sqrt{14 x-x^2} d(x-7) \\ = & (x-7) \sqrt{14 x-x^2}-\int(x-7) \frac{14-2 x}{2 \sqrt{14 x-x^2}} d x \\ = & (x-7) \sqrt{14 x-x^2}-\int(x-7) \frac{7-x}{\sqrt{14 x-x^2}} d x \\ = & (x-7) \sqrt{14 x-x^2}-\int \frac{14 x-x^2-49}{\sqrt{14 x-x^2}} d x \\ = & (x-7) \sqrt{14 x-x^2}-I+49 \int \frac{d x}{\sqrt{14 x-x^2}} \\ \end{aligned} $$ Rearranging yields $$\int \sqrt{14 x-x^2} d x = \frac{1}{2}(x-7) \sqrt{14 x-x^2}+\frac{49}{2} \underbrace{\int \frac{d x}{\sqrt{49-(x-7)^2}}}_J$$ For the integral $J$, $$ \begin{aligned} \int \frac{1}{\sqrt{14 x-x^2}} d x = & \int \frac{1}{\sqrt{x} \sqrt{14-x}} dx\\ = & -2 \int \frac{1}{\sqrt{14-(\sqrt{14-x})^2}} d(\sqrt{14-x}) \\ = & -2 \sin ^{-1}\left(\frac{\sqrt{14-x}}{\sqrt{14}}\right) +C\\ = & -2 \sin ^{-1}\left(\sqrt{1-\frac{x}{14}}\right)+C \end{aligned} $$ Plugging back gives WA’s answer $$\int \sqrt{14 x-x^2} d x = \frac{1}{2}(x-7) \sqrt{14 x-x^2} -49 \sin ^{-1} \sqrt{1-\frac{x}{14}}+C $$