Integral problem?

Question

int (root(4 - x^2)) = ?

I know the answer involves arcsin somehow but have no idea how to get there, im a bit rusty at these and would appreciate the method used as well as the answer

Answer 1

Are you sure its just the square root or is it 1 over the square root of that number?

Answer 2

you're right. Using x=2sin u ,dx=2cos udu. puttiing into the equation we get
int (root(4-[2sin u]^2)*2cos udu
4int(cos u)^2 du using the double angle formula we get
(cos u)^2 =[1+cos 2u]/2 then our equation becomes
2int(1+cos 2u)du
2(u+0.5sin2u)+C
2[arcsin(x/2) +{sin(2arcsin(x/2))}/2] +C

Answer 3

int root(4-x^2)
we can write it as

(2^2-x^2) ok
now from formula

root(a^2-x^2)= x/2 root(a^2-x^2)+a^2/2sin inverse x/a

put th values in that u will get the ans

Answer 4

integral of(4-x^2)^1/2dx.
put x=2sin u, dx=2cosu du
integral becomes:
integral of [2cosu]2cosu du
2 integral of [cos 2u +1] du
=2*[1/2 sin 2u+u]+C
=sin2u+2u
=2sinu cosu +2u+C
=2*x/2 root(1-x^2/4)+2arcsin(x/2)]+C
=x/2 root(4-x^2)+2arcsiin(x/2).+CANS.

Answer 5

substitute x for 2sin z . so we get dx=2cos z dz
put the value in the integral, we get int(2cos z dz/2cos z.)=int dz=z+c=sin^-1(x/2)+c

Answer 6

= (x/2)(sqrt(4-x^2)) +2arcsin (x/2) + C