Start from the given
∫2+x√4−2x−x2dx
Start with Algebra by completing the square
4−2x−x2=−(x2+2x−4)=−(x2+2x+1−1−4)
and
4−2x−x2=−((x+1)2−5)=5−(x+1)2
then
∫2+x√4−2x−x2dx=∫2+x√5−(x+1)2dx
The Trigonometric Substitution
Let x+1=√5⋅sinθ
and x=√5⋅sinθ−1
and dx=√5⋅cosdθ
Let's do the substitution
∫2+x√5−(x+1)2dx=
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5−(√5⋅sinθ)2
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5−5⋅sin2θ
continue simplification by trigonometric identities
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅√1−sin2θ
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅√cos2θ
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅cosθ
∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅cosθ
and
∫2+x√5−(x+1)2dx=∫(√5sinθ+1)dθ
∫2+x√5−(x+1)2dx=∫(1+√5sinθ)dθ
∫2+x√5−(x+1)2dx=θ−√5⋅cosθ+C
Now, time to imagine your right triangle with
angle θ
Let x+1 the Opposite side to angle θ
Let √5 the Hypotenuse
Let √4−2x−x2 the Adjacent side to angle θ
Return the variables
∫2+x√5−(x+1)2dx=θ−√5⋅cosθ+C
∫2+x√5−(x+1)2dx=arcsin(x+1√5)−√4−2x−x2+C
I hope the explanation is useful....God bless...
