Answer
$\dfrac{1}{4 \sqrt 3}$
Work Step by Step
Let us consider $x =2 \sin \theta $
and $dx=2 \cos \theta d \theta $ and $2 \cos \theta =\sqrt{4-x^2}$
Now, the given integral becomes (without limits):
$\int \dfrac{2 \cos \theta d \theta }{8 \cos ^3 \theta} =\dfrac{1}{4} \int \sec^2 \theta d\theta$
Now, integrate with limits and plug in $\tan \theta =\sin\theta/\cos \theta=\dfrac{x}{\sqrt{4-x^2}}$:
$\dfrac{1}{4} \int\tan \theta +C=(\dfrac{1}{4})[\dfrac{x}{(4-x^2)^{1/2}}]_0^{1}=(\dfrac{1}{4})[\dfrac{1}{(4-1)^{1/2}}-0]=\dfrac{1}{4 \sqrt 3}$
