<s>[INST] 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?
 [/INST] 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.
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