What is the integral of sec^3(x)?

The integral of

\sec^3(x)

\frac{1}{2} \sec(x) \tan(x) + \frac{1}{2}\ln \lvert \sec(x) + \tan(x) + C

.

Solution. We want to determine the integral of

\sec^3(x)

, i.e.:

\begin{align*} I = \int \sec^3(x) dx. \end{align*}

Notice that we assign the integral of

\sec^3(x)

I

, where we see later why we do this on purpose. We will now integrate by parts, so we will use the following formula:

\begin{align*} \int UdV = UV - \int VdU, \end{align*}

where we have the following equations:

\begin{align*} U = \sec(x), \quad &dV = \sec^2(x)dx\\ dU = \sec(x)\tan(x)dx, \quad &V = \tan(x). \end{align*}

You can verify yourself here for

dU

, and here for

V

. So we get the following:

\begin{align*} I &= \int \sec^3(x) dx \\ &= \sec(x)\tan(x) - \int \sec(x)\tan^2(x)dx. \end{align*}

We have seen here that

\tan^2(x) = \sec^2(x) - 1

. So we get:

\begin{align*} \sec(x)\tan(x) - \int \sec(x)\tan^2(x) &= \sec(x)\tan(x) - \int \sec(x)(\sec^3(x) - 1)dx \\ &= \sec(x)\tan(x) - \int \sec^3(x)dx + \int \sec(x)dx \\ &= \sec(x)\tan(x) - I + \ln \lvert \sec(x) + \tan(x) \rvert, \end{align*}

where we have seen here the integral of

\sec(x)

. Now we will bring

I

to the left-hand side, and so, we get the integral of

\sec^3(x)

\begin{align*} I &= sec(x)\tan(x) - I + \ln \lvert \sec(x) + \tan(x) \rvert \iff \\ 2I &= \sec(x)\tan(x) + \ln \lvert \sec(x) + \tan(x) \rvert \iff \\ \int \sec^3(x)dx &= \frac{1}{2} \sec(x)\tan(x) + \frac{1}{2}\ln \lvert \sec(x) + \tan(x) \rvert + C \end{align*}