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