Use trigonometric substitution. We have an integral with a square root of the form √a2−x2. Therefore, use the substitution x=asinθ.
In our case, a=2. Our substitution will therefore be x=2sinθ. Then dx−2cosθdθ.
⇒∫√4−(2sinθ)2(2sinθ)2⋅2cosθdθ
⇒∫√4−4sin2θ4sin2θ⋅2cosθdθ
⇒∫√4(1−sin2θ)4sin2θ⋅2cosθdθ
Use sin2x+cos2x=1:
⇒∫√4cos2θ4sin2θ⋅2cosθdθ
⇒∫2cosθ4sin2θ⋅2cosθdθ
⇒∫4cos2θ4sin2θdθ
⇒∫cos2θsin2θdθ
Use cotx=1tanx=1sinxcosx=cosxsinx:
⇒∫cot2θdθ
Use cot2x+1=csc2x:
⇒∫csc2θ−1dθ
⇒∫csc2θdθ−∫1dθ
These are both trivial integrals.
⇒−cotθ−θ+C
We know from our original substitution that x2=sinθ. We can deduce that θ=arcsin(x2). Therefore, the side adjacent θ is √4−x2 in measure. Thus, cotθ=√4−x2x.
⇒−√4−x2x−arcsin(x2)+C
Hopefully this helps!
