\[\begin{align*} \boxed{\int \csc(x) \, dx = \ln\left| \tan\left( \dfrac{x}{2} \right) \right| + C} \end{align*}\]
where \( C \) is the constant of integration.
Introduction:
In this article, we will explore how to find the integral of the cosecant function, \( \csc(x) \). The integral of \( \csc(x) \) is less straightforward than those of \( \sin(x) \) or \( \cos(x) \), but with the right technique, it can be evaluated easily. To make the integral more manageable, we can manipulate the integrand by multiplying the numerator and denominator by \( \csc(x) + \cot(x) \):
\[\begin{align*} \int \csc(x) \, dx = \int \csc(x) \cdot \dfrac{ \csc(x) + \cot(x) }{ \csc(x) + \cot(x) } \, dx \end{align*}\]
Notice the integrand is not changing its value because we are multiplying by 1.
\[\begin{align*} \int \dfrac{ \csc^2(x) + \csc(x)\cot(x) }{ \csc(x) + \cot(x) } \, dx \end{align*}\]
Recognize the Derivative in the Numerator
Notice that the derivative of the denominator \( \csc(x) + \cot(x) \) is:
\[\begin{align*} \dfrac{d}{dx} [ \csc(x) + \cot(x) ] = -\csc(x)\cot(x) – \csc^2(x) = – [ \csc^2(x) + \csc(x)\cot(x) ] \end{align*}\]
This means:
\[\begin{align*} – \dfrac{d}{dx} [ \csc(x) + \cot(x) ] = \csc^2(x) + \csc(x)\cot(x) \end{align*}\]
This is exactly our numerator! Substituting into the integral:
\[\begin{align*} \int \dfrac{ \csc^2(x) + \csc(x)\cot(x) }{ \csc(x) + \cot(x) } \, dx = -\int \dfrac{ d [ \csc(x) + \cot(x) ] }{ \csc(x) + \cot(x) } dx \end{align*}\]
\[\begin{align*} – \int \dfrac{ d [ \csc(x) + \cot(x) ] }{ \csc(x) + \cot(x) } = -\ln\left| \csc(x) + \cot(x) \right| + C \end{align*}\]
Using trigonometric identities, we can express the result in terms of \( \tan\left( \dfrac{x}{2} \right) \):
\[\begin{align*} – \ln\left| \csc(x) + \cot(x) \right| = \ln\left| \tan\left( \dfrac{x}{2} \right) \right| + C \end{align*}\]
Therefore, the integral is:
\[\begin{align*} \boxed{\int \csc(x) \, dx = \ln\left| \tan\left( \dfrac{x}{2} \right) \right| + C} \end{align*}\]
Note that if the $\csc x + \cot x$ are a little confusing to work with, you could always set $u = \csc x + \cot x$ to get $\displaystyle \int \csc x = \int \dfrac{1}{u} \, du$, and proceed from there as a common integral.
