what is the integration of square root 4-x2?

Answer 1

By x2, I'm going to assume you mean x^2.

Integral ( sqrt(4 - x^2) dx )

You definitely have to use trigonometric substitution to solve this. I'm going to use "t" instead of "theta" though.

Let x = 2sin(t). Then
dx = 2cos(t) dt

Applying the substitution gives us

Integral ( sqrt(4 - (2sin(t))^2) 2cos(t) dt )

Simplifying,

Integral ( sqrt(4 - 4sin^2(t)) 2cos(t) dt )

Factor the 4 inside the square root, and the 2 outside of the integral.

2 * Integral ( sqrt [ 4(1 - sin^2(t)) ] cos(t) dt )

The 4 comes out of the square root as 2.

2 * Integral ( 2 * sqrt(1 - sin^2(t)) cos(t) dt )

Pull the 2 out of the integral, as it is a constant.

2 * 2 * Integral ( sqrt(1 - sin^2(t)) cos(t) dt )

Multiply the 2's together. Use the trig identity 1 - sin^2(t) = cos^2(t).

4 * Integral ( sqrt( cos^2(t) ) cos(t) dt )

The square root of a squared term is itself.

4 * Integral ( cos(t) cos(t) dt )

4 * Integral ( cos^2(t) dt )

To integrate squared trigonometric functions by themselves, we must use the half angle identity. A reminder that

cos^2(t) = (1/2) (1 + cos(2t))

4 * Integral ( (1/2) (1 + cos(2t)) dt )

Factor the (1/2) out of the integral. This merged with the 4 to become 2.

2 * Integral ( (1 + cos(2t)) dt )

And now, we can integrate term by term. A reminder that, for a constant n, the integral of cos(nt) = (1/n)sin(nt).

2 (t + (1/2) sin(2t) ) + C

Distribute the 2,

2t + sin(2t) + C

Apply the double angle identity sin(2t) = 2sin(t)cos(t)

2t + 2sin(t)cos(t) + C

To convert this back in terms of x, we must use trigonometry and SOHCAHTOA. Since

x = 2sin(t), then
sin(t) = x/2

A reminder that by SOHCAHTOA,
sin(t) = opp/hyp. Therefore,

opp = x,
hyp = 2, so by Pythagoras,
adj = sqrt(2^2 - x^2) = sqrt(4 - x^2)

It follows that
cos(t) = adj/hyp = sqrt(4 - x^2)/2

Furthermore, since sin(t) = x/2, t = arcsin(x/2).
Using all of these facts,

2t + 2sin(t)cos(t) + C

becomes

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

Which converts to

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

Answer 2

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RE:
what is the integration of square root 4-x2?

Answer 3

Sqrt 4-x 2

Answer 4

I = ∫ ( 2 ² - x ² )^(1/2) dx is a standard integral found in text books.
I = sin ^(-1) (x/2) + C

Answer 5

∫ √(4-x^2)

=1/2 √(4-x^2) + 2arcsin(x/2) + C


arc sin----> sin^-1 ...sine inverse

Answer 6

negative 16 ? Is this a Algebra Q ? or what is the square root of negative 4 ?