Evaluate the integral
\[ \int \csc^3(x) \; dx \]
Write the integrand \( \csc^3(x) \) as the product \( \csc x \csc^2 x \)
\[ \int \csc^3(x) \; dx = \int \csc x \csc^2 x ; dx \]
Use the integration by parts given by: \( \int u' v \; dx = u v - \int u v' \; dx \).
Let \( u' = \csc^2 x \) , \( v = \csc x \) and hence \( u = - \cot x \) , \( v' = -\csc x \cot x \)
and the integral may be written as
\[ \int \csc^3(x) \; dx = (-\cot x)(\csc x) - \int ((- \cot x)(-\csc x \cot x)) \; dx\]
Simplify the above
\[ \int \csc^3(x) \; dx = - \cot x \csc x - \int \cot^2 x \csc x \; dx\]
Use the trigonometric identity \( \cot^2 x = \csc^2 x - 1 \) and write the integral as follows
\[ \int \csc^3(x) \; dx = - \cot x \csc x - \int (\csc^2 x - 1) \csc x \; dx\]
Expand the integrand of the integral \( \displaystyle \int (\csc^2 x - 1) \csc x \; dx \)
\[ \int \csc^3(x) \; dx = - \cot x \csc x - \int \csc^3 x \; dx + \int \csc x \; dx\]
Use the common integral \( \displaystyle \int \csc x \; dx = \ln|\csc x - \cot x| \) and rewrite the integral as
\[ \int \csc^3(x) \; dx = - \cot x \csc x - \int \csc^3 x \; dx + \ln|\csc x - \cot x| \]
Add \( \displaystyle \int \csc^3 x \; dx \) to both sides of the above and simplify to obtain
\[ 2 \int \csc^3(x) \; dx = - \cot x \csc x + \ln|\csc x - \cot x| \]
Multiply all terms by \( \dfrac{1}{2} \) and simplify to obtain the final answer as
\[ \boxed { \int \csc^3(x) \; dx = - \dfrac{1}{2} \cot x \csc x + \dfrac{1}{2} \ln|\csc x - \cot x| + c }\]
