Integration of the secant function

by Evan Pfeuffer

Let’s start with integration of the $\sec(x)$ :

$\int \sec(x) \,dx$

We can express this as

$\int \frac{1}{\cos(x)} \,dx$

If we multiply the integrand by $\frac{\cos(x)}{\cos(x)}$ then we obtain:

$\int \frac{\cos(x)}{\cos^2(x)} \, dx$

Using our trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ we can convert the integrand into:

$\int \frac{\cos(x)}{1-\sin^2(x)} \,dx$

This allows us to make the substitution $u = \sin(x)$ , $du = \cos(x) \, dx$

Continuing with the substitution:

$\int \frac{\cos(x)}{1-\sin^2(x)} \,dx = \int \frac{1}{1-u^2} \,du$

Next we need to continue with a partial fraction decomposition of $\frac{1}{1-u^2} = \frac{1}{(1-u)(1+u)}$ :

Let $A$ and $B$ be constants such that:

$\frac{A}{1-u} + \frac{B}{1+u} = \frac{1}{(1-u)(1+u)}$

Cross-multiplying and setting the numerators equals yields:

$A(1+u) + B(1-u) = 1$

Letting $u = -1$ implies that $A(0) + 2B = 1$ or $B = 1/2$

Letting $u = 1$ implies that $A(2) + B(0) = 1$ or $A = 1/2$

Now we can restate our integral as:

$\int \frac{1}{1-u^2} \,du = \int \frac{1/2}{1-u} + \frac{1/2}{1+u} \,du = \frac{1}{2} \int \frac{1}{1-u} + \frac{1}{1+u} \,du$

Using the fact that $\int \frac{1}{x} \,dx = \ln |x| + C$

We can integrate as follows:

$\frac{1}{2} \int \frac{1}{1-u} + \frac{1}{1+u} \,du = -\frac{1}{2} \ln|1-u| + \frac{1}{2}\ln|1+u| + C$

$-\frac{1}{2} \ln|1-u| + \frac{1}{2}\ln|1+u| + C = \frac{1}{2} \ln |\frac{1+u}{1-u}| + C$

Remembering our original substitution $u = \sin(x)$ we finally have:

$\int \sec(x) \,dx = \frac{1}{2} \ln |\frac{1+\sin(x)}{1-\sin(x)}| + C$

Another common result for $\int \sec(x) \,dx$ is $\ln|\sec(x) + \tan(x)| + C$

Here’s how they are equivalent:

$\frac{1}{2} \ln |\frac{1+\sin(x)}{1-\sin(x)}| + C = \ln \sqrt{\frac{1+\sin(x)}{1-\sin(x)}}+ C$

$\ln \sqrt{\frac{1+\sin(x)}{1-\sin(x)}}+ C = \ln \sqrt{\frac{1+\sin(x)}{1-\sin(x)} \times \frac{1+\sin(x)}{1+\sin(x)}}+ C = \ln \sqrt{\frac{(1+\sin(x))^2}{1-\sin^2(x)}} +C$

$\ln \sqrt{\frac{(1+\sin(x))^2}{1-\sin^2(x)}} +C = \ln \sqrt{\frac{(1+\sin(x))^2}{\cos^2(x)}} +C = \ln \bigl| \frac{1+\sin(x)}{\cos(x)} \bigr| + C$

And finally,

$\ln \bigl| \frac{1+\sin(x)}{\cos(x)} \bigr| + C = \ln \bigl| \sec(x) + \tan(x) \bigr| + C$

Calculus is Cool

Integration of the secant function

One response to “Integration of the secant function”

Leave a comment Cancel reply

Navigation

About

Integration of the secant function

Share this:

One response to “Integration of the secant function”

Leave a comment Cancel reply

Navigation

About