Finding $a>1$ such that $\int_0^\pi \tfrac{\sin x\,\mathrm{d}x} { \sqrt{1 − 2a\cos x + a^2}}$ equals given constant $a > 1$, solve for $a$:
$$\int_0^\pi \frac{\sin(x)} { (1 − 2a\cos(x) + a^2)^{1/2}} dx = 0.2018$$
I have attempted to use substitution $u = -\cos x$, $\frac{du}{dx} = \sin x$, limits $[1, -1]$. 
I end up with $\frac{1}{a}(1 + 2a + a^2)^{1/2} - \frac{1}{a}(1 - 2a + a^2)^{1/2}  = 0.2018$. 
I believe this is not the right direction. I would greatly appreciate any ideas on how to solve this. 
 A: Let us consider your integral:
$$I=\int_0^\pi \frac{\sin(x)} { (1 − 2a\cos(x) + a^2)^{1/2}}dx,\;\ a>1.$$
Consider a triangle $ABC$ with angles $\alpha,\beta,\gamma$ opposite to the sides $a,b,c$, respectively:

The law of cosines states that
$$c^2=a^2+b^2-2ab\cos(\gamma).$$
To keep with this notation, let $b=1$ and $\gamma=x$.
Our integral becomes:
$$I=\int_0^\pi \frac{\sin(\gamma)} {c}d\gamma.$$
The law of sines states that
$$\frac{\sin(\alpha)}{a}=\frac{\sin(\gamma)}{c}.$$
Our integral becomes:
$$I=\int_0^\pi \frac{\sin(\alpha)}{a}d\gamma.$$
The three angles $\alpha,\beta,\gamma$ add up to $\pi$ radians.
Let
$$\gamma=\pi-\alpha-\beta,$$
such that
$$d\gamma=-d\alpha-d\beta.$$
When 
$$\gamma\rightarrow 0,\;\ \alpha\rightarrow \pi,\;\ \beta\rightarrow 0.$$
When
$$\gamma\rightarrow \pi,\;\ \alpha\rightarrow 0,\;\ \beta\rightarrow 0.$$
Our integral becomes:
$$I=\int_\pi^0 \frac{\sin(\alpha)}{a}(-d\alpha)+\int_0^0 \frac{\sin(\alpha)}{a}(-d\beta).$$
Regarding the integral with respect to $\alpha$, incorporate the a minus sign into the integral in order to switch the interval. Recognize that the integral with respect to $\beta$ is equal to zero:
$$I=\int_0^\pi \frac{\sin(\alpha)}{a}d\alpha.$$
Integration yields:
$$I=-\frac{\cos(\alpha)}{a}\bigg |_0^\pi.$$
Evaluating yields:
$$I=\frac{2}{a}.$$
It follows that
$$\frac{2}{a}=0.2018.$$
Therefore,
$$a=\frac{2}{0.2018}=\frac{10000}{1009}.$$
A: If the term in the denominator is $a^{2}$ not $x^{2}$, Your answer appears to be correct. Just go further:
$$\frac{\lvert{a+1}\rvert}{a}-\frac{\lvert{a-1}\rvert}{a}=0.2018$$
since $a>1$:
$$\frac{2}{a}=0.2018\\
a = \frac{2}{0.2018}$$
