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!

Learn how to evaluate the integral of $$ \tan^3x $$ with a step-by-step solution, applying trigonometric identities and substitution techniques. This problem covers advanced trigonometric integration concepts suitable for students in Grades 11-12 or first-year calculus.

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