integration of tan cube x

The integration of tan cube x $\tan ^3 x$, can be found using integration substitution and trigonometry identities . The integral of $\tan ^3 x$ with respect to (x) is:

\[
\int \tan^3 x \, dx =\frac {1}{2} \tan^2 x – \ln |sec x|+ C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof of integration of tan cube x

To integrate $\tan^3(x)$, we can use trigonometric identities and integration techniques. The integral in question is:

$$
\int \tan^3(x) \, dx
$$

Let’s break $\tan^3(x)$ into $\tan^2(x) \tan(x)$ and then use the trigonometric identity $\tan^2(x) = \sec^2(x) – 1$ to rewrite the integral. This gives us:

$$
\int (\sec^2(x) – 1) \tan(x) \, dx
$$

This can be split into two integrals:

$$
\int \sec^2(x) \tan(x) \, dx – \int \tan(x) \, dx
$$

The first integral can be solved by a simple substitution, and the second integral is a standard integral, the value is
$\int \tan(x) \, dx= \ln |sec x|$

Let’s calculate it.

$$
\int \sec^2(x) \tan(x) \, dx
$$
let t = tan x
$dt = \sec^2(x) dx$
Therefore

$$
=\int t \, dt= \frac {t^2}{2}
$$

Therefore,
The integral $\int \tan^3(x) \, dx$ simplifies to:

$$
\frac {1}{2} \tan^2 x – \ln |sec x|+ C$$

where $C$ is the constant of integration. This result combines the effects of integrating $\tan(x)$ with the adjusted form of $\tan^3(x)$ using the identity $\tan^2(x) = \sec^2(x) – 1$.