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:
\[ 8(\alpha + \beta) = 3. \]
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: