The Indefinite Integrals of Secant and Cosecant

# The Indefinite Integrals of Secant and Cosecant

Evaluating $\int \sec x \: dx$ and $\int \csc x \: dx$ may seem difficult at first, but in actuality, it is not that difficult if we apply a special trick for these integrals.

Theorem 1: The following functions have the following indefinite integrals:a) If $f(x) = \sec x$, then $\int \sec x \: dx = \int \sec x \: dx = \ln \mid \sec x + \tan x \mid + C$.b) If $f(x) = \csc x \: dx$, then $\int \csc x \: dx = - \ln \mid \csc x + \cot x \mid + C$. |

**Proof of a):**We are first going to start by multiplying the integrand by $\frac{\sec x \tan x}{\sec x \tan x }$, which may seem odd at first, but puts us in a much more appropriate situation for integration:

\begin{align} \int \sec x \cdot \frac{\sec x \tan x}{\sec x \tan x } \: dx \\ \int \frac{\sec ^2x + \sec x \tan x}{\sec x \tan x} \: dx \end{align}

- Now let's make a substitution. Let $u = \sec x + \tan x$. It thus follows that $du = \sec x \tan x + \sec ^2 x \: dx$. We thus get:

\begin{align} \int \frac{1}{u} \: du \\ \ln | u | + C \end{align}

- Which we can now resubstitute back in $u$ in terms of $x$ to get:

\begin{align} \int \sec x \: dx = \ln | \sec x + \tan x | + C \quad \blacksquare \end{align}

**Proof of b):**First let's multiply the integrand by $\frac{\csc x + \cot x}{\csc x + \cot x}$. We thus get:

\begin{align} \int \csc x \cdot \frac{\csc x + \cot x}{\csc x + \cot x} \: dx \\ \int \frac{ \csc ^2 x + \cot x}{\csc x + \cot x} \: dx \end{align}

- Now let's make the substitution $u = \csc x + \cot x$. It thus follows that $du = -\csc x \cot x - \csc ^2 x \: dx$, or more appropriately, $- (\csc x \cot x + \csc ^2 x$. Hence we get that:

\begin{align} \int - \frac{1}{u} \: du \\ - \ln | u | + C \end{align}

- Hence by substituting back $u$ in terms of $x$, we obtain that:

\begin{align} \: \int \csc x \: dx = - \ln \mid \csc x + \cot x \mid + C \quad \blacksquare \end{align}