can you show me how to maniplate this using Trig subtitution...

integral dx/4+x^2 x is bounded on [-2, 2]

Sure! To manipulate this integral using trigonometric substitution, we need to find an appropriate substitution that will simplify the integrand.

Let's start by making the substitution x = 2tan(t). This substitution will help us simplify the expression 4 + x^2.

First, let's find dx in terms of dt:
dx = 2sec^2(t) dt

Now, let's substitute x and dx in the integral:
∫ dx / (4 + x^2) = ∫ (2sec^2(t) dt) / (4 + (2tan(t))^2)

Simplifying the denominator:
= ∫ (2sec^2(t) dt) / (4 + 4tan^2(t))
= ∫ (2sec^2(t) dt) / 4(1 + tan^2(t))
= ∫ (2sec^2(t) dt) / 4sec^2(t)
= (1/2) ∫ (dt / sec^2(t))

Now, we can simplify the integrand further:
= (1/2) ∫ dt / (1/cos^2(t))
= (1/2) ∫ cos^2(t) dt

At this point, let's convert cos^2(t) into a form that's easier to integrate. Using the identity sin^2(t) + cos^2(t) = 1, we can rearrange it as cos^2(t) = 1 - sin^2(t).

Substituting back:
= (1/2) ∫ (1 - sin^2(t)) dt

Now, we can integrate term by term:
= (1/2) (t - (1/3)sin^3(t)) + C

Finally, we need to revert back to the original variable, x:
= (1/2) [arctan(x/2) - (1/3)(x/2)√(4 + x^2)] + C

To evaluate the integral over the interval [-2, 2], substitute the upper and lower limits into the antiderivative equation and subtract the result of the lower limit from the upper limit:

[(1/2) [arctan(2/2) - (1/3)(2/2)√(4 + 2^2)] - (1/2) [arctan(-2/2) - (1/3)(-2/2)√(4 + (-2)^2)]

Simplifying further if needed.