Integration of sec x

Integration of sec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of sec x are

I

\[
\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C
\]

II

\[
\int \sec(x) \, dx = -\ln | \sec(x) – \tan(x) | + C
\]

III

\[
\int \csc(x) \, dx = \ln | \tan ( \frac {x}{2} + \frac {\pi}{2})| + C
\]

IV

\[
\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {1 + \sin x}{1 – \sin x} | + C
\]

Lets check out the proof of each of these

Proof of I

Integral of secant can be little tricky and can be solved by multiplying and dividing by $(\sec(x) + \tan(x))$. Here’s how it’s done:

\[
\int \sec(x) \, dx = \int \sec(x) \cdot \frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} \, dx \\
= \int \frac{\sec^2(x) + \sec(x)\tan(x)}{\sec(x) + \tan(x)} \, dx.
\]

Now, taking u as
$ u= \sec(x) + \tan(x) $
$ du =(\sec(x)\tan(x) + \sec^2(x)) dx $

Therefore integration becomes

\[
\int \sec(x) \, dx =\int \frac {1}{u} \, du,
\]

hence

\[
\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C
\]

Proof of II

We can write the result of I as

\[
\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C = \ln | \frac {\sec^2(x) – \tan^2(x)}{\sin(x) – \sin(x) } | + C \\
=\ln | \frac {1}{\sec(x) – \tan(x) } | + C = \ln | (\sec(x) – \tan(x))^{-1} | + C \\
=- \ln | \sec(x) – \tan(x) | + C
\]

Proof of III

This can be provided using trigonometric formulas

\[
\int \sec(x) \, dx =\int \frac {1}{\cos(x)} \, dx
=\int \frac {1}{\sin(x + \frac {\pi}{2})} \, dx
=\int \frac {1}{2 \sin(x/2 + \pi/4) \cos(x/2 + \pi/4)} \, dx
\]

Dividing and Multiplying $cos(x/2 + \pi/4)$ in the denominator

\[
\int \sec(x) \, dx = \frac {1}{2}\int \frac {\sec^2(x/2+ \pi/4)}{\tan(x/2+ \pi/4)} \, dx
\]

Lets $u=\tan(x/2 + \pi/4)$
$du= \frac{1}{2} \sec^2(x/2 + \pi/4) dx$

Therefore

\[
\int \sec(x) \, dx = \int \frac {1}{u} \, du = ln |u| + C
\]

Hence

\[
\int \sec(x) \, dx = \ln | \tan (\frac {x}{2} +\frac {\pi}{2}) | + C
\]

Proof of IV

This can be proof by using integration by partial fractions

\[
\int \sec(x) \, dx =\int \frac {1}{\cos(x)} \, dx
=\int \frac {cos(x)}{\cos^2(x)} \, dx
=\int \frac {cos(x)}{ 1- \sin^2(x)} \, dx
\]

Lets $ u=\sin(x)$
$du= \cos(x) dx$

Therefore

\[
\int \sec(x) \, dx =\int \frac {1}{1 -u^2} \, du
=\frac {1}{2} \int [\frac {1}{1+u} + \frac {1}{1-u} ]\, du
=\frac {1}{2} ln |\frac {1+u}{1-u}| + C
\]

Hence

\[
\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {1 + \sin x}{ 1- \sin x} | + C
\]

Solved Example

Example 1

\[
\int_{0}^{\pi/4} \sec(x) \, dx.
\]

Solution

Using the result from above:

\[
\int_{0}^{\pi/4} \sec(x) \, dx = \left[ \ln|\sec(x) + \tan(x)| \right]_{0}^{\pi/4}.
\]

Evaluating at the bounds:

\[
\begin{align} \left[ \ln|\sec(x) + \tan(x)| \right]_{0}^{\pi/4}
&= \ln|\sec(\pi/4) + \tan(\pi/4)| – \ln|\sec(0) + \tan(0)| \\
&= \ln|(\sqrt{2}) + 1| – \ln|1 + 0| \\
&= \ln(\sqrt{2} + 1). \end{align}
\]

So, $\int_{0}^{\pi/4} \sec(x) \, dx = \ln(\sqrt{2} + 1)$.