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

The integral of

\tan^3(x)

\frac{1}{2}\tan^2(x) + \ln(\lvert \cos(x) \rvert) + C

.

Solution. We want determine the integral of

\tan^3(x)

, i.e.,

\begin{align*} \int \tan^3(x) dx. \end{align*}

We can simplify the integral of

\tan^3(x)

to the following:

\begin{align*} \int \tan^3(x) dx = \int (\sec^2(x) - 1)\tan(x)dx = \int \sec^2(x)\tan(x)dx - \int \tan(x)dx. \end{align*}

where we have seen here that

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

. Further, we know from here the integral of

\tan(x)

\ln \lvert \sec(x) \rvert

plus some constant. So we get:

\begin{align*} \int \sec^2(x)\tan(x)dx - \int \tan(x)dx = \int \sec^2(x)\tan(x)dx - \ln \lvert \sec(x) \rvert = \int \sec^2(x)\tan(x)dx + \ln( \lvert \cos(x) \rvert). \end{align*}

To do the other integral, we will apply the substitution method. Let

u = \tan(x)

, then we know from here that

du = \sec^2(x)dx

. Therefore, combining everything, we get:

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

Therefore, the integral of

\tan^3(x)

\frac{1}{2}\tan^2(x) + \ln(\lvert \cos(x) \rvert) + C