Determine the following integrals: 2.1 ∫ a³ˣ sin2x dx 2.2 ∫ (3 / cot⁴ 2x) dx 2.3 ∫ sin⁵ ax cos³ ax dx 2.4 ∫ (1 / (3x² - 2x + 9)) dx

Step 1: Solve integral 2.1.
We need to evaluate $\int a^{3x} \sin(2x) dx$.
This integral requires integration by parts twice. We can use the formula for $\int e^{kx} \sin(mx) dx = \frac{e^{kx}}{k^2+m^2}(k \sin(mx) - m \cos(mx)) + C$.
First, rewrite $a^{3x}$ as $e^{3x \ln a}$. So, $k = 3 \ln a$ and $m = 2$.

$$\int a^{3x} \sin(2x) dx = \int e^{3x \ln a} \sin(2x) dx$$

Applying the formula:

$$= \frac{e^{3x \ln a}}{(3 \ln a)^2 + 2^2} (3 \ln a \sin(2x) - 2 \cos(2x)) + C$$

Substitute $e^{3x \ln a}$ back to $a^{3x}$:

$$= \frac{a^{3x}}{(3 \ln a)^2 + 4} (3 \ln a \sin(2x) - 2 \cos(2x)) + C$$ $$= \boxed{\frac{a^{3x}}{9 (\ln a)^2 + 4} (3 \ln a \sin(2x) - 2 \cos(2x)) + C}$$

Step 2: Solve integral 2.2.
We need to evaluate $\int \frac{3}{\cot^4(2x)} dx$.
Use the identity $\frac{1}{\cot x} = \tan x$.

$$\int \frac{3}{\cot^4(2x)} dx = \int 3 \tan^4(2x) dx$$

Let $u = 2x$, so $du = 2 dx$, which means $dx = \frac{1}{2} du$.

$$= 3 \int \tan^4 u \left(\frac{1}{2}\right) du = \frac{3}{2} \int \tan^4 u du$$

Use the reduction formula for $\int \tan^n u du = \frac{\tan^{n-1} u}{n-1} - \int \tan^{n-2} u du$.
For $n=4$:

$$\int \tan^4 u du = \frac{\tan^3 u}{3} - \int \tan^2 u du$$

For $\int \tan^2 u du$, use the identity $\tan^2 u = \sec^2 u - 1$:

$$\int \tan^2 u du = \int (\sec^2 u - 1) du = \tan u - u + C_1$$

Substitute this back:

$$\int \tan^4 u du = \frac{\tan^3 u}{3} - (\tan u - u) + C_1$$

Now substitute this back into the main integral:

$$\frac{3}{2} \left( \frac{\tan^3 u}{3} - \tan u + u \right) + C$$

Substitute $u = 2x$:

$$= \frac{3}{2} \left( \frac{\tan^3(2x)}{3} - \tan(2x) + 2x \right) + C$$ $$= \frac{1}{2} \tan^3(2x) - \frac{3}{2} \tan(2x) + 3x + C$$ $$= \boxed{\frac{1}{2} \tan^3(2x) - \frac{3}{2} \tan(2x) + 3x + C}$$

Step 3: Solve integral 2.3.
We need to evaluate $\int \sin^5(ax) \cos^3(ax) dx$.
Since the power of $\cos(ax)$ is odd, let $u = \sin(ax)$.
Then $du = a \cos(ax) dx$, so $\cos(ax) dx = \frac{1}{a} du$.
Rewrite $\cos^3(ax)$ as $\cos^2(ax) \cos(ax) = (1 - \sin^2(ax)) \cos(ax)$.

$$\int \sin^5(ax) \cos^3(ax) dx = \int \sin^5(ax) (1 - \sin^2(ax)) \cos(ax) dx$$

Substitute $u = \sin(ax)$:

$$= \int u^5 (1 - u^2) \frac{1}{a} du$$ $$= \frac{1}{a} \int (u^5 - u^7) du$$

Integrate term by term:

$$= \frac{1}{a} \left( \frac{u^6}{6} - \frac{u^8}{8} \right) + C$$

Substitute back $u = \sin(ax)$:

$$= \frac{1}{a} \left( \frac{\sin^6(ax)}{6} - \frac{\sin^8(ax)}{8} \right) + C$$ $$= \boxed{\frac{\sin^6(ax)}{6a} - \frac{\sin^8(ax)}{8a} + C}$$

Step 4: Solve integral 2.4.
We need to evaluate $\int \frac{1}{3x^2 - 2x + 9} dx$.
Complete the square in the denominator.

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