Answer
$\int_{1}^{2} arctan~x~dx~~$ has the largest value since $~~arctan~x \gt arctan \sqrt{x} \gt arctan(sin~x)~~$ on the interval $~~1 \lt x \leq 2$
Work Step by Step
On the interval $1 \lt x \leq 2$:
$x \gt \sqrt{x} \gt sin~x$
$arctan~x \gt arctan \sqrt{x} \gt arctan(sin~x)$
Therefore:
$\int_{1}^{2} arctan~x~dx \gt \int_{1}^{2} arctan\sqrt{x} \gt \int_{1}^{2} arctan(sin~x)$
$\int_{1}^{2} arctan~x~dx$ has the largest value since $arctan~x \gt arctan \sqrt{x} \gt arctan(sin~x)$ on the interval $1 \lt x \leq 2$
