Answer
$-\dfrac{2\sqrt{4-w^2}}{w}+C$
Work Step by Step
Let us consider $w =2 \sin \theta $
and $dw=2 \cos \theta d\theta $ and $2 \cos \theta =\sqrt{4-w^2}$
Now, the given integral becomes:
$\int \dfrac{(16\cos \theta)}{4 \sin^2 \theta} d \theta = 2 \int \csc^2 \theta d \theta$
Integrate and plug in $\cot \theta =\cos\theta/\sin \theta=\dfrac{\sqrt{4-w^2}}{w}$
Thus, $-2\cot \theta+C=-\dfrac{2\sqrt{4-w^2}}{w}+C$
