U-sub or by parts?

$\begin{aligned} I & = \int_{-1}^1 \sin (\arccos x) \ dx & \small \color{#3D99F6} \text{Let }\cos \theta = x, \ -\sin \theta \ d \theta = dx \\ & = \int_0 ^\pi \sin^2 \theta \ d \theta \\ & = \int_0 ^\pi \frac {1-\cos 2 \theta}2 \ d \theta \\ & = \frac \theta 2 - \frac {\sin 2 \theta}4 \ \bigg|_0 ^\pi \\ & = \boxed{\dfrac \pi 2} \end{aligned}$

$\sin(\arccos x) = \sqrt{1- \cos^2 (\arccos x)} = \sqrt{1-x^2 }$ on the given interval.

Use trigonometric substitution or realise that this is a semicircle and the answer is $\dfrac{\pi}{2}$ , and in the form asked for, $1+2 = \boxed{3}$

You get $\int ^ { 1 } _ { 1 } \sin ( \arccos x ) dx = \frac { \pi } { 2 }$ by following this site. Auto calculator

So, the answer is $1 + 2 = \boxed { 3 }$ .