Integral of sin and cos invers

$\large \displaystyle I = \int_0^{\frac14} \frac{ \arccos(\sqrt{x}) - \arcsin(\sqrt{x}) } { \arccos(\sqrt{x}) + \arcsin(\sqrt{x}) } dx$

$\color{#20A900} \boxed{\large \displaystyle u = \arcsin(\sqrt{x})} \rightarrow \boxed{\large \displaystyle x = \sin^2(u)} \rightarrow \boxed{\large \displaystyle \frac{\pi}{2} - u = \arccos(\sqrt{x})}$

$\color{#20A900} \large \displaystyle du = \frac{1}{\sqrt{1 - x}} \frac{1}{2\sqrt{x}} dx \rightarrow dx = 2 \sqrt{1 - \sin^2(u)}\sqrt{\sin^2(u)} du \rightarrow \boxed{\large \displaystyle dx = \sin(2u)du }$

$\large \displaystyle I = \int_0^{\frac{\pi}{6}} \frac{ \frac{\pi}{2} - 2u}{ \frac{\pi}{2}} \sin(2u) du$

$\large \displaystyle I = \int_0^{\frac{\pi}{6}} \sin(2u) du - \frac{4}{\pi} \int_0^{\frac{\pi}{6}} u \sin(2u) du$

$\large \displaystyle I = -\frac{\cos(2u)}{2} \Bigg |_0^{\frac{\pi}{6}} - \frac{4}{\pi} \left ( - \frac{u \cos(2u)}{2} \Bigg |_0^{\frac{\pi}{6}} + \int_0^{\frac{\pi}{6}} \frac{\cos(2u)}{2}du \right )$

$\large \displaystyle I = \frac14 - \frac{4}{\pi} \left ( \frac{\pi}{24} + \frac{\sin(2u)}{4} \Bigg |_0^{\frac{\pi}{6}} \right )$

$\large \displaystyle I = \frac14 - \frac{4}{\pi} \left ( \frac{\pi}{24} + \frac{\sqrt{3}}{8} \right )$

$\large \displaystyle I = \frac14 + \frac16 - \frac{\sqrt{3}}{2 \pi}$

$\color{#20A900} \boxed{ \large \displaystyle I = \frac{5}{12} - \frac{\sqrt{3}}{2 \pi} }$

$\color{#3D99F6} \large \displaystyle a = 5, b = 12, c = 3, d = 2, \boxed { \large \displaystyle a + b + c + d = 22}$