Evaluating Definite Integral $\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx$ How can I prove that

$$\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3})$$

Can someone help me, please?
 A: Let $\arcsin\left(\dfrac{4-3\sqrt{x^2-1}}{5x}\right) = t$. We then have
$$\dfrac{4-3\sqrt{x^2-1}}{5x} = \sin(t)\implies(4-5x \sin(t))^2 = 9(x^2-1)$$
This gives us
$$25x^2 \sin^2(t)-9x^2 - 40x \sin(t) + 25 = 0 \implies (4x \sin(t)-5)^2 = (3x \cos(t))^2$$
Hence, now let us set
$$x = \dfrac5{3\cos(t) + 4 \sin(t)} = \sec\left(t+\phi\right)$$
where $\cos(\phi) = \dfrac35$ and $\sin(\phi) = -\dfrac45$.
Hence,
$$\int tdx = tx - \int xdt$$
I trust you can now plug in the appropriate limits for $t$ and obtain the answer.
A: Using Trigonometric substitution
let $x=\sec\theta$
$\displaystyle\implies (i) 0\le \theta\le\pi$ using the definition of principal value 
and  $\displaystyle(ii)\frac{4-3\sqrt{x^2-1}}{5x}=\frac{4-3|\tan\theta|}{5\sec\theta}$
Method $1:$ If $\displaystyle0\le \theta\le\frac\pi2,$ 
when $x=2, \theta=\text{arcsec}2=\frac\pi3$
when $x=1, \theta=\text{arcsec}1=0$
$\displaystyle\frac{4-3\sqrt{x^2-1}}{5x}=\frac45\cos\theta-\frac35\sin\theta=\sin\left(\arcsin\frac45-\theta\right)$ (using this)
$$\implies\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\int_0^{\frac\pi3}\left(\arcsin\frac45-\theta\right)\sec\theta\cdot\tan\theta d\theta$$
Now integrating by parts, 
$$\int \left(\arcsin\frac45-\theta\right)\sec\theta\cdot\tan\theta d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\int\sec\theta\cdot\tan\theta d\theta-\int\left(\frac{d(\arcsin\frac45-\theta)}{d\theta}\cdot\int\sec\theta\cdot\tan\theta d\theta\right)d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\sec\theta-\int\left(-1\cdot\sec\theta\right)d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\sec\theta+\ln|\sec\theta+\tan\theta|+C$$
Observe that  $\displaystyle 2\text{arcsec}(\sqrt5) =2\arccos\frac1{\sqrt5}=\cos\left(\frac25-1\right)$
$\displaystyle=\cos\left(-\frac35\right)=\pi-\arccos\frac35=\pi-\arcsin\frac45$  as $\arccos x=+\arcsin\sqrt{1-x^2}\forall $ real $x$ based on the Principal value of inverse cosine ratio
Method $2:$ 
If $\displaystyle \frac\pi2\le \theta\le\pi,\text{arcsec}(2)=?$
$$\frac{4-3\sqrt{x^2-1}}{5x}=\frac45\cos\theta+\frac35\sin\theta=\sin\left(\arcsin\frac45+\theta\right)$$
A: I found an answer from the solution of my another question that is here.
$$\begin{align}\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)\,dx&=\int_1^2\arcsin\left(\frac{\sqrt{x^2-1}+2+2-4\sqrt{x^2-1}}{\sqrt{5}\sqrt{5}x}\right)\,dx\\
&=\int_0^{\pi/3}\arcsin(\sin(2\text{arcsec}(\sqrt{5})+\alpha))\tan(\alpha)\sec(\alpha)\,d\alpha\\
&=\int_0^{\pi/3}(\pi-2\text{arcsec}(\sqrt{5})-\alpha)\tan(\alpha)\sec(\alpha)\,d\alpha\\
&=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3}).\end{align}$$
