Integral of \( \csc^3(x) \)

Derivation and Evaluation

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 integration by parts given by: \( \int u' v \, dx = uv - \int u v' \, dx \).

Let \( u' = \csc^2 x \) and \( v = \csc x \), which gives \( u = -\cot x \) and \( v' = -\csc x \cot x \).

Substituting into the integration by parts formula gives:

\[ \int \csc^3(x) \, dx = (-\cot x)(\csc x) - \int (-\cot x)(-\csc x \cot x) \, dx \]

Simplify the expression:

\[ \int \csc^3(x) \, dx = -\cot x \csc x - \int \cot^2 x \csc x \, dx \]

Use the trigonometric identity \( \cot^2 x = \csc^2 x - 1 \) to rewrite the integral:

\[ \int \csc^3(x) \, dx = -\cot x \csc x - \int (\csc^2 x - 1) \csc x \, dx \]

Expand the integrand:

\[ \int \csc^3(x) \, dx = -\cot x \csc x - \int \csc^3 x \, dx + \int \csc x \, dx \]

Use the standard integral \( \displaystyle \int \csc x \, dx = \ln|\csc x - \cot x| \) and substitute it into the expression:

\[ \int \csc^3(x) \, dx = -\cot x \csc x - \int \csc^3 x \, dx + \ln|\csc x - \cot x| \]

Add \( \displaystyle \int \csc^3 x \, dx \) to both sides and simplify:

\[ 2 \int \csc^3(x) \, dx = -\cot x \csc x + \ln|\csc x - \cot x| \]

Multiplying all terms by \( \dfrac{1}{2} \) and adding the constant of integration \( c \), we obtain the final answer: