# Need help in evaluation of indefinite integral. y= square root (-2x^2+4x+6)

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We'll factorize by -2 the radicand:

-2x^2+4x+6 = -2(x^2 - 2x - 3)

We'll re-write the factor x^2 - 2x - 3 = x^2 - 2x + 1 - 4

We notice that we've get the perfect square (x - 1)^2

x^2 - 2x + 1 - 4 = (x - 1)^2 - 4

Int f(x)dx = sqrt2 Int sqrt[4 - (x - 1)^2]dx

We'll substitute x- 1 = t.

We'll differentiate both sides:

dx = dt

sqrt2 Int sqrt[4 - (t)^2]dt

We'll substitute t = 2 sin u

u = arcsin (t/2)

We'll differentiate both sides:

dt = 2cos udu

sqrt2 Int sqrt[4 - 4(sin u)^2]*2cos udu = 4sqrt2 Int (cos u)^2du

4sqrt2 Int (cos u)^2du = (4sqrt2/2) Int (1 + cos 2u)du

(4sqrt2/2) Int (1 + cos 2u)du = (4sqrt2/2) [Int du + Int (cos 2u)du]

Int f(x)dx = (2*u*sqrt2) + sqrt 2*sin 2u + C

**Int f(x)dx = 2sqrt2*arcsin [(x-1)/2] +sqrt 2*sin {2arcsin [(x-1)/2]} + C**