Start from the given
int (2+x)/sqrt(4-2x-x^2)dx∫2+x√4−2x−x2dx
Start with Algebra by completing the square
4-2x-x^2=-(x^2+2x-4)=-(x^2+2x+1-1-4)4−2x−x2=−(x2+2x−4)=−(x2+2x+1−1−4)
and
4-2x-x^2=-((x+1)^2-5)=5-(x+1)^24−2x−x2=−((x+1)2−5)=5−(x+1)2
then
int (2+x)/sqrt(4-2x-x^2)dx=int (2+x)/sqrt(5-(x+1)^2)dx∫2+x√4−2x−x2dx=∫2+x√5−(x+1)2dx
The Trigonometric Substitution
Let x+1=sqrt(5)*sin thetax+1=√5⋅sinθ
and x=sqrt(5)*sin theta -1x=√5⋅sinθ−1
and dx=sqrt(5)*cos d thetadx=√5⋅cosdθ
Let's do the substitution
int (2+x)/sqrt(5-(x+1)^2)dx=∫2+x√5−(x+1)2dx=
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-(sqrt(5)*sin theta)^2)∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5−(√5⋅sinθ)2
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-5*sin^2 theta)∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5−5⋅sin2θ
continue simplification by trigonometric identities
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(1-sin^2 theta))∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅√1−sin2θ
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(cos^2 theta))∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅√cos2θ
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅cosθ
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)∫(√5sinθ+1)⋅(√5⋅cosθ)dθ√5⋅cosθ
and
int (2+x)/sqrt(5-(x+1)^2)dx=int(sqrt(5)sin theta + 1)d theta∫2+x√5−(x+1)2dx=∫(√5sinθ+1)dθ
int (2+x)/sqrt(5-(x+1)^2)dx=int(1+sqrt(5)sin theta )d theta∫2+x√5−(x+1)2dx=∫(1+√5sinθ)dθ
int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C∫2+x√5−(x+1)2dx=θ−√5⋅cosθ+C
Now, time to imagine your right triangle with
angle thetaθ
Let x+1x+1 the Opposite side to angle thetaθ
Let sqrt5√5 the Hypotenuse
Let sqrt(4-2x-x^2)√4−2x−x2 the Adjacent side to angle thetaθ
Return the variables
int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C∫2+x√5−(x+1)2dx=θ−√5⋅cosθ+C
int (2+x)/sqrt(5-(x+1)^2)dx=arcsin ((x+1)/(sqrt5))-sqrt(4-2x-x^2)+C∫2+x√5−(x+1)2dx=arcsin(x+1√5)−√4−2x−x2+C
I hope the explanation is useful....God bless...
