Integral of \( \csc x \)

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Evaluate the integral \[ \int \csc x \; dx \]
Note that 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 as
\[ \int \csc x \; dx = \int \dfrac{1}{\sin x} \; dx \]
\( \sin x \) may be written as \[ \sin x = \sin (2 (\frac{x}{2}) ) \]
Use
trigonometric identity \( \sin (2x) = 2 \sin x \cos x \) to write \[ \sin x = 2 \sin \left(\dfrac{x}{2} \right) \cos \left(\dfrac{x}{2} \right) \] and substitute in 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 \( u = \sin \left(\dfrac{x}{2} \right) \) and hence \( \dfrac{du}{dx} = \dfrac{1}{2} \cos \left(\dfrac{x}{2} \right) \) or \( dx = \dfrac{2}{\cos \left(\dfrac{x}{2} \right)} du \) and substitute in the above integral to obtain \[ \int \csc x \; dx = \int \dfrac{1}{2 u \cos \left(\dfrac{x}{2} \right)} \dfrac{2}{\cos \left(\dfrac{x}{2} \right)} du \] Simplify \[ \int \csc x \; dx = \int \dfrac{1}{ u \cos^2 \left(\dfrac{x}{2} \right)} du \]
Use
trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) to write \[ \cos^2 \left(\dfrac{x}{2} \right) = 1 - \sin^2 \left(\dfrac{x}{2} \right) = 1 - u^2 \] and substitute in the above integral to obtain \[ \int \csc x \; dx = \int \dfrac{1}{ u (1-u^2)} du \] Use partial fractions decomposition to write the integrand as \[ \dfrac{1}{ u (1-u^2)} = \dfrac{1}{u}-\dfrac{1}{2\left(u+1\right)}-\dfrac{1}{2\left(u-1\right)} \] Substitute in the integral \[ \int \csc x \; dx = \int \left(\dfrac{1}{u}-\dfrac{1}{2\left(u+1\right)}-\dfrac{1}{2\left(u-1\right)}\right) du \\ = \int \dfrac{1}{u} du - \int \dfrac{1}{2\left(u+1\right)} du - \int \dfrac{1}{2\left(u-1\right)} du \] Use the formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to obtain \[ \int \csc x \; dx = \ln |u| - \dfrac{1}{2} \ln |u+1| - \dfrac{1}{2} \ln |u-1| + c\] Group the logarithmic expressions using the properties \( \quad \ln \dfrac{a}{b \cdot c} = \ln a - \ln b - \ln c \) and \( \quad \ln \dfrac{1}{2} a = \ln \sqrt {a} \) to write \[ \int \csc x \; dx = \ln |u| - \left(\dfrac{1}{2} \ln |u+1| + \dfrac{1}{2} \ln |u-1| \right) + c \\ = \ln \dfrac{|u|}{\sqrt {|u+1| |u - 1|} } + c \\ = \ln \dfrac{|u|}{\sqrt {|u^2 - 1|} } + c \] Note that \( |u^2 - 1| = |1 - u^2| \) and hence \[ \int \csc x \; dx = = \ln \dfrac{|u|}{\sqrt {|1 - u^2|} } + c\] Substitute back \( u = \sin \left(\dfrac{x}{2} \right) \) \[ \int \csc x \; dx = \ln \dfrac{\left|\sin \left(\dfrac{x}{2} \right)\right|}{\sqrt { \left|1 - \left(\sin \left(\dfrac{x}{2} \right)\right)^2 \right|} } +c \] Use trigonometric identity \( \cos^2 \left(\dfrac{x}{2} \right) = 1 - \sin^2 \left(\dfrac{x}{2} \right) \) to calculate the above integral \[ \int \csc x \; dx = \ln \dfrac{\left|\sin \left(\dfrac{x}{2} \right)\right|}{\sqrt {\left| \left(\cos \left(\dfrac{x}{2} \right)\right)^2\right|} } +c \\ \] Simplify using \( \sqrt {\left|(\cos \left(\dfrac{x}{2} \right))^2\right|} = \left|\cos \left(\dfrac{x}{2} \right)\right| \) \[ \int \csc x \; dx = \ln \dfrac{\left|\sin \left(\dfrac{x}{2} \right)\right|}{\left|\cos \left(\dfrac{x}{2} \right)\right|} +c \] Use trigonometric identity \( \tan x = \dfrac{\sin x }{\cos x } \) and the property of the absolute value \( \dfrac{|a|}{|b|} = \left|\dfrac{a}{b}\right|\) to write the final answer as \[ \boxed { \int \csc x \; dx = \ln \left|\tan \left(\dfrac{x}{2} \right)\right| +c } \]



More References and Links

  1. Table of Integral Formulas
  2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
  3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
  4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8