Integral of \( \csc x \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \csc x \, dx \]

Note: There are several methods to calculate the above integral. The steps described below allow one to review many ideas in trigonometry and in calculating integrals; also the final result is quite simple.

Use \( \csc x = \dfrac{1}{\sin x} \) to rewrite the integral:

\[ \int \csc x \, dx = \int \dfrac{1}{\sin x} \, dx \]

\( \sin x \) may be written using a double-angle formula as:

\[ \sin x = \sin\left(2 \cdot \dfrac{x}{2}\right) \]

Use the trigonometric identity \( \sin(2x) = 2 \sin x \cos x \) to write:

\[ \sin x = 2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right) \]

Substitute this into the integral:

\[ \int \csc x \, dx = \int \dfrac{1}{2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right)} \, dx \]

Use Integration by Substitution: let \( u = \sin\left(\dfrac{x}{2}\right) \), which gives \( \dfrac{du}{dx} = \dfrac{1}{2} \cos\left(\dfrac{x}{2}\right) \) or \( dx = \dfrac{2}{\cos\left(\dfrac{x}{2}\right)} \, du \). Substituting into the integral:

\[ \int \csc x \, dx = \int \dfrac{1}{2 u \cos\left(\dfrac{x}{2}\right)} \cdot \dfrac{2}{\cos\left(\dfrac{x}{2}\right)} \, du \]

Simplify the expression:

\[ \int \csc x \, dx = \int \dfrac{1}{u \cos^2\left(\dfrac{x}{2}\right)} \, du \]

Use the trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) to write:

\[ \cos^2\left(\dfrac{x}{2}\right) = 1 - \sin^2\left(\dfrac{x}{2}\right) = 1 - u^2 \]

Substitute into the integral:

\[ \int \csc x \, dx = \int \dfrac{1}{u(1 - u^2)} \, du \]

Use partial fraction decomposition to write the integrand as:

\[ \dfrac{1}{u(1 - u^2)} = \dfrac{1}{u} - \dfrac{1}{2(u + 1)} - \dfrac{1}{2(u - 1)} \]

Substitute back into the integral and split:

\[ \begin{aligned} \int \csc x \, dx &= \int \left(\dfrac{1}{u} - \dfrac{1}{2(u + 1)} - \dfrac{1}{2(u - 1)}\right) du \\[6pt] &= \int \dfrac{1}{u} \, du - \int \dfrac{1}{2(u + 1)} \, du - \int \dfrac{1}{2(u - 1)} \, du \end{aligned} \]

Use the integration formula \( \displaystyle \int \dfrac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c \):

\[ \int \csc x \, dx = \ln|u| - \dfrac{1}{2} \ln|u + 1| - \dfrac{1}{2} \ln|u - 1| + c \]

Group the logarithmic expressions using properties of logarithms (\( \ln a - \ln b - \ln c = \ln\frac{a}{bc} \) and \( \frac{1}{2}\ln a = \ln\sqrt{a} \)):

\[ \begin{aligned} \int \csc x \, dx &= \ln|u| - \left(\dfrac{1}{2} \ln|u + 1| + \dfrac{1}{2} \ln|u - 1|\right) + c \\[6pt] &= \ln|u| - \ln\sqrt{|u + 1||u - 1|} + c \\[6pt] &= \ln\left(\dfrac{|u|}{\sqrt{|u^2 - 1|}}\right) + c \end{aligned} \]

Note that \( |u^2 - 1| = |1 - u^2| \), hence:

\[ \int \csc x \, dx = \ln\left(\dfrac{|u|}{\sqrt{|1 - u^2|}}\right) + c \]

Substitute back \( u = \sin\left(\dfrac{x}{2}\right) \):

\[ \int \csc x \, dx = \ln\left(\dfrac{\left|\sin\left(\dfrac{x}{2}\right)\right|}{\sqrt{\left|1 - \left(\sin\left(\dfrac{x}{2}\right)\right)^2\right|}}\right) + c \]

Use the trigonometric identity \( \cos^2\left(\dfrac{x}{2}\right) = 1 - \sin^2\left(\dfrac{x}{2}\right) \):

\[ \int \csc x \, dx = \ln\left(\dfrac{\left|\sin\left(\dfrac{x}{2}\right)\right|}{\sqrt{\left|\left(\cos\left(\dfrac{x}{2}\right)\right)^2\right|}}\right) + c \]

Simplify using \( \sqrt{\left|\left(\cos\left(\dfrac{x}{2}\right)\right)^2\right|} = \left|\cos\left(\dfrac{x}{2}\right)\right| \):

\[ \int \csc x \, dx = \ln\left(\dfrac{\left|\sin\left(\dfrac{x}{2}\right)\right|}{\left|\cos\left(\dfrac{x}{2}\right)\right|}\right) + c \]

Use the trigonometric identity \( \tan x = \dfrac{\sin x}{\cos x} \) and absolute value properties to write the final answer: