1. Find $\int \frac{1}{\sqrt{9-4x^2}} \, dx.$ 2 - HSC - SSCE Mathematics Extension 2 - Question 1 - 2007 - Paper 1

Using the identity ( \tan^2 x = \sec^2 x - 1 ), we can simplify the integral:

$\int \tan^2 x \sec^2 x \, dx = \int (\sec^2 x - 1) \sec^2 x \, dx = \int \sec^4 x \, dx - \int \sec^2 x \, dx.$

The integral of ( \sec^2 x ) is ( \tan x ) and requires techniques such as reduction for ( \int \sec^4 x , dx ). The solution can be derived from integration by parts or known formulas.

We will use integration by parts, where we let ( u = x ) and ( dv = \cos x , dx ight) Then ( du = dx ) and ( v = \sin x ).

Using integration by parts, we get:

$\int_0^1 x \cos x \, dx = [x \sin x]_0^1 - \int_0^1 \sin x \, dx = [\sin x]_0^1 = \sin 1 - 0 = \sin 1.$

Let ( u = 1 - x \Rightarrow du = -dx ), and then the limits change, transforming the integral into:

$\int_1^{-2} \frac{1-u}{\sqrt{u}} (-du) = \int_{-2}^1 \left( \frac{1}{\sqrt{u}} - \sqrt{u} \right) du.$

After appropriate limits and simplifications, evaluate each part to find the final result.

Consider the identity provided. We can find:

$\int_2^2 \frac{2}{x^3+x^2+x+1} \, dx = \int_2^2 \left( \frac{1}{x^2 + 1} - \frac{x}{x^2+x+1} \right) dx = 0,$

as the integral over a zero-length interval is simply zero.