If $$\int \csc^5 x \, dx = \alpha \cot x \csc x \left(\csc^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$$ where $$\alpha, \beta \in \mathbb{R}$$ and $$C$$ is the constant of integration, then the value of $$8(\alpha + \beta)$$ equals _____

To solve the integral $$\int \csc^5 x dx$$, we use integration by parts and reduction formulas. Recall that $$\csc x = \frac{1}{\sin x}$$ and $$\cot x = \frac{\cos x}{\sin x}$$.

Set $$I_5 = \int \csc^5 x dx$$. Express $$\csc^5 x = \csc^3 x \cdot \csc^2 x$$. For integration by parts, let:

$$u = \csc^3 x, \quad dv = \csc^2 x dx$$

Then,

$$du = -3 \csc^3 x \cot x dx, \quad v = \int \csc^2 x dx = -\cot x$$

Using the integration by parts formula $$\int u dv = uv - \int v du$$:

$$I_5 = \csc^3 x \cdot (-\cot x) - \int (-\cot x) \cdot (-3 \csc^3 x \cot x) dx$$

Simplify:

$$I_5 = -\csc^3 x \cot x - 3 \int \cot^2 x \csc^3 x dx$$

Substitute $$\cot^2 x = \csc^2 x - 1$$:

$$I_5 = -\csc^3 x \cot x - 3 \int (\csc^2 x - 1) \csc^3 x dx$$

$$I_5 = -\csc^3 x \cot x - 3 \int \csc^5 x dx + 3 \int \csc^3 x dx$$

Let $$I_3 = \int \csc^3 x dx$$, so:

$$I_5 = -\csc^3 x \cot x - 3 I_5 + 3 I_3$$

Solve for $$I_5$$:

$$I_5 + 3 I_5 = -\csc^3 x \cot x + 3 I_3$$

$$4 I_5 = -\csc^3 x \cot x + 3 I_3$$

$$I_5 = \frac{1}{4} \left( -\csc^3 x \cot x + 3 I_3 \right) \quad \text{(1)}$$

Now, compute $$I_3 = \int \csc^3 x dx$$. Use integration by parts with:

$$u = \csc x, \quad dv = \csc^2 x dx$$

Then,

$$du = -\csc x \cot x dx, \quad v = -\cot x$$

So,

$$I_3 = \csc x \cdot (-\cot x) - \int (-\cot x) \cdot (-\csc x \cot x) dx$$

$$I_3 = -\csc x \cot x - \int \csc x \cot^2 x dx$$

Substitute $$\cot^2 x = \csc^2 x - 1$$:

$$I_3 = -\csc x \cot x - \int \csc x (\csc^2 x - 1) dx$$

$$I_3 = -\csc x \cot x - \int \csc^3 x dx + \int \csc x dx$$

$$I_3 = -\csc x \cot x - I_3 + \int \csc x dx$$

Solve for $$I_3$$:

$$I_3 + I_3 = -\csc x \cot x + \int \csc x dx$$

$$2 I_3 = -\csc x \cot x + \int \csc x dx \quad \text{(2)}$$

The integral $$\int \csc x dx$$ is a standard result:

$$\int \csc x dx = \ln \left| \tan \frac{x}{2} \right| + C$$

Substitute into equation (2):

$$2 I_3 = -\csc x \cot x + \ln \left| \tan \frac{x}{2} \right| + C_1$$

$$I_3 = \frac{1}{2} \left( -\csc x \cot x + \ln \left| \tan \frac{x}{2} \right| \right) + C_2 \quad \text{(3)}$$

Substitute equation (3) into equation (1):

$$I_5 = \frac{1}{4} \left[ -\csc^3 x \cot x + 3 \cdot \frac{1}{2} \left( -\csc x \cot x + \ln \left| \tan \frac{x}{2} \right| \right) \right] + C$$

$$I_5 = \frac{1}{4} \left[ -\csc^3 x \cot x + \frac{3}{2} \left( -\csc x \cot x + \ln \left| \tan \frac{x}{2} \right| \right) \right] + C$$

Distribute $$\frac{3}{2}$$:

$$I_5 = \frac{1}{4} \left[ -\csc^3 x \cot x - \frac{3}{2} \csc x \cot x + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right| \right] + C$$

Factor $$\csc x \cot x$$:

$$I_5 = \frac{1}{4} \left[ \csc x \cot x \left( -\csc^2 x - \frac{3}{2} \right) + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right| \right] + C$$

Note that $$-\csc^2 x - \frac{3}{2} = - \left( \csc^2 x + \frac{3}{2} \right)$$, so:

$$I_5 = \frac{1}{4} \left[ -\csc x \cot x \left( \csc^2 x + \frac{3}{2} \right) + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right| \right] + C$$

Distribute $$\frac{1}{4}$$:

$$I_5 = -\frac{1}{4} \csc x \cot x \left( \csc^2 x + \frac{3}{2} \right) + \frac{1}{4} \cdot \frac{3}{2} \ln \left| \tan \frac{x}{2} \right| + C$$

$$I_5 = -\frac{1}{4} \csc x \cot x \left( \csc^2 x + \frac{3}{2} \right) + \frac{3}{8} \ln \left| \tan \frac{x}{2} \right| + C$$

Compare to the given form:

$$\int \csc^5 x dx = \alpha \cot x \csc x \left(\csc^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$$

Note that $$\cot x \csc x = \csc x \cot x$$. Equating coefficients:

For $$\csc x \cot x \left( \csc^2 x + \frac{3}{2} \right)$$, coefficient is $$-\frac{1}{4}$$, so $$\alpha = -\frac{1}{4}$$.

For $$\ln \left| \tan \frac{x}{2} \right|$$, coefficient is $$\frac{3}{8}$$, so $$\beta = \frac{3}{8}$$.

Now compute $$\alpha + \beta$$:

$$\alpha + \beta = -\frac{1}{4} + \frac{3}{8} = -\frac{2}{8} + \frac{3}{8} = \frac{1}{8}$$

Then,

$$8(\alpha + \beta) = 8 \times \frac{1}{8} = 1$$

Thus, the value is 1.