To evaluate the integral \( \int \csc^5 x \, dx \), we use integration by parts. Let
\[ I = \int \csc^3 x \cdot \csc^2 x \, dx. \]
Applying integration by parts, we let:
\[ I = -\cot x \csc^3 x + \int \cot x \cdot (-3 \csc^2 x \cot x \csc x) \, dx. \]
Simplifying, we get:
\[ I = -\cot x \csc^3 x - 3 \int \csc^3 x (\csc^2 x - 1) \, dx, \] \[ I = -\cot x \csc^3 x - 3I + 3 \int \csc^3 x \, dx. \]
Let
\[ I_1 = \int \csc^3 x \, dx = -\csc x \cot x - \int \cot^2 x \csc x \, dx. \]
Using this and simplifying further, we identify values for \( \alpha \) and \( \beta \). After solving, we find:
\[ 8(\alpha + \beta) = 3. \]
Let $ f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 $ and $ 2g(x) - 3g\left( \frac{1}{2} \right) = x, \, x>0. \, \text{If} \, \alpha = \int_{1}^{2} f(x) \, dx, \, \beta = \int_{1}^{2} g(x) \, dx, \text{ then the value of } 9\alpha + \beta \text{ is:}$
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)