To solve the integral \( \int \tan^3(2x) \sec(2x) \, dx \), we will break it down step by step.
### Step 1: Rewrite the Integral
We can rewrite \( \tan^3(2x) \) as \( \tan^2(2x) \tan(2x) \). This allows us to express \( \tan^2(2x) \) in terms of \( \sec(2x) \):
\[
\tan^2(2x) = \sec^2(2x) - 1
\]
Thus, we can rewrite the integral as:
\[
\int \tan^3(2x) \sec(2x) \, dx = \int \tan^2(2x) \tan(2x) \sec(2x) \, dx = \int (\sec^2(2x) - 1) \tan(2x) \sec(2x) \, dx
\]
### Step 2: Distribute the Integral
Now we can distribute the integral:
\[
\int (\sec^2(2x) - 1) \tan(2x) \sec(2x) \, dx = \int \sec^3(2x) \tan(2x) \, dx - \int \tan(2x) \sec(2x) \, dx
\]
### Step 3: Use Substitution
For the first integral \( \int \sec^3(2x) \tan(2x) \, dx \), we will use the substitution:
\[
t = \sec(2x) \quad \text{then} \quad dt = 2\sec(2x)\tan(2x) \, dx \quad \Rightarrow \quad dx = \frac{dt}{2\sec(2x)\tan(2x)}
\]
Thus, we can rewrite the integral:
\[
\int \sec^3(2x) \tan(2x) \, dx = \int t^3 \cdot \frac{dt}{2t \tan(2x)} = \frac{1}{2} \int t^2 \, dt
\]
### Step 4: Integrate
Now we can integrate:
\[
\frac{1}{2} \int t^2 \, dt = \frac{1}{2} \cdot \frac{t^3}{3} = \frac{1}{6} t^3 = \frac{1}{6} \sec^3(2x)
\]
### Step 5: Solve the Second Integral
Now we solve the second integral:
\[
-\int \tan(2x) \sec(2x) \, dx
\]
Using the same substitution \( t = \sec(2x) \):
\[
-\int \tan(2x) \sec(2x) \, dx = -\frac{1}{2} \int dt = -\frac{1}{2} t = -\frac{1}{2} \sec(2x)
\]
### Step 6: Combine Results
Combining both results, we have:
\[
\int \tan^3(2x) \sec(2x) \, dx = \frac{1}{6} \sec^3(2x) - \frac{1}{2} \sec(2x) + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \tan^3(2x) \sec(2x) \, dx = \frac{1}{6} \sec^3(2x) - \frac{1}{2} \sec(2x) + C
\]
