To evaluate the integral $\int \tan^3x \, \mathrm{d}x$, we can start by rewriting $\tan^3x$ as $\tan x \cdot \tan^2x$, and then express $\tan^2x$ in terms of $\sec^2x$ using the identity:

$\tan^2x = \sec^2x - 1$

Thus, we have:

$\tan^3x = \tan x (\sec^2x - 1)$

Now, the integral becomes:

$\int \tan^3x \, \mathrm{d}x = \int \tan x (\sec^2x - 1) \, \mathrm{d}x$

We can split this into two integrals:

$\int \tan x (\sec^2x - 1) \, \mathrm{d}x = \int \tan x \sec^2x \, \mathrm{d}x - \int \tan x \, \mathrm{d}x$

First Integral: $\int \tan x \sec^2x \, \mathrm{d}x$

For this integral, use substitution: Let $u = \tan x$, so $\frac{du}{dx} = \sec^2x$.

This simplifies the first integral to:

$\int u \, \mathrm{d}u = \frac{u^2}{2} = \frac{\tan^2x}{2}$

Second Integral: $\int \tan x \, \mathrm{d}x$

The integral of $\tan x$ is:

$\int \tan x \, \mathrm{d}x = \ln|\sec x| + C$

Combining the Results

Now, combining both parts, we get:

$\int \tan^3x \, \mathrm{d}x = \frac{\tan^2x}{2} - \ln|\sec x| + C$

Thus, the final answer is:

$\boxed{\frac{\tan^2x}{2} - \ln|\sec x| + C}$

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Would you like a breakdown of any part or additional details?

Here are 5 related questions:

1. How do you integrate $\int \tan^2x \, \mathrm{d}x$?
2. How do you handle the integral $\int \sec^3x \, \mathrm{d}x$?
3. Can you solve $\int \cot^3x \, \mathrm{d}x$ using a similar method?
4. What is the general method for integrating powers of trigonometric functions?
5. Can substitution always be used for integrating trigonometric expressions?

Tip: Memorizing key trigonometric identities (like $\tan^2x = \sec^2x - 1$) can simplify many integration problems!