# Integral of $\csc x$

 

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 $\boxed { \int \csc x \; dx = \ln \left|\tan \left(\dfrac{x}{2} \right)\right| +c }$