The value of `inttan^(3)2xsec2x` dx is equal to:

Options

A. `1/3sec^(2)2x-1/2sec2x+C`
B. `-1/6sec^(3)2x-1/2sec2x+C`
C. `1/6sec^(3)2x-1/2sec2x+C`
D. `1/3sec^(3)2x+1/2sec2x+C`

Answer

To solve the integral \( \int \tan^3(2x) \sec(2x) \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Let:
\[ I = \int \tan^3(2x) \sec(2x) \, dx \]
We can express \( \tan^3(2x) \) as \( \tan^2(2x) \tan(2x) \) and use the identity \( \tan^2(2x) = \sec^2(2x) - 1 \):
\[ I = \int \tan^2(2x) \sec(2x) \tan(2x) \, dx = \int (\sec^2(2x) - 1) \sec(2x) \tan(2x) \, dx \]

### Step 2: Use Substitution
Let:
\[ t = \sec(2x) \]
Then, we differentiate \( t \):
\[ \frac{dt}{dx} = 2 \sec(2x) \tan(2x) \implies dt = 2 \sec(2x) \tan(2x) \, dx \implies dx = \frac{dt}{2 \sec(2x) \tan(2x)} \]
Substituting this into the integral gives:
\[ I = \int (\sec^2(2x) - 1) t \cdot \frac{dt}{2t} = \frac{1}{2} \int (t^2 - 1) \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ I = \frac{1}{2} \left( \frac{t^3}{3} - t \right) + C \]
Substituting back \( t = \sec(2x) \):
\[ I = \frac{1}{2} \left( \frac{\sec^3(2x)}{3} - \sec(2x) \right) + C \]

### Step 4: Simplify
This simplifies to:
\[ I = \frac{1}{6} \sec^3(2x) - \frac{1}{2} \sec(2x) + C \]

### Final Result
Thus, the value of the integral is:
\[ \int \tan^3(2x) \sec(2x) \, dx = \frac{1}{6} \sec^3(2x) - \frac{1}{2} \sec(2x) + C \]