`int_(0)^(pi//8) tan^(2) 2x dx` is equal to

Options

A. `(4- pi )/( 8 )`
B. `( 4+ pi )/( 8)`
C. `( 4- pi )/( 4)`
D. `( 4- pi )/( 2)`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We know that:
\[ \tan^2(2x) = \sec^2(2x) - 1 \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx = \int_{0}^{\frac{\pi}{8}} (\sec^2(2x) - 1) \, dx \]

### Step 2: Separate the integral
Now, we can separate the integral into two parts:
\[ \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx = \int_{0}^{\frac{\pi}{8}} \sec^2(2x) \, dx - \int_{0}^{\frac{\pi}{8}} 1 \, dx \]

### Step 3: Solve the first integral
The integral of \( \sec^2(2x) \) can be computed using the formula:
\[ \int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + C \]
For our case, \( k = 2 \):
\[ \int \sec^2(2x) \, dx = \frac{1}{2} \tan(2x) \]
Thus, we evaluate:
\[ \int_{0}^{\frac{\pi}{8}} \sec^2(2x) \, dx = \left[ \frac{1}{2} \tan(2x) \right]_{0}^{\frac{\pi}{8}} \]
Calculating the limits:
\[ = \frac{1}{2} \tan\left(2 \cdot \frac{\pi}{8}\right) - \frac{1}{2} \tan(0) = \frac{1}{2} \tan\left(\frac{\pi}{4}\right) - 0 = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Step 4: Solve the second integral
The integral of 1 from 0 to \( \frac{\pi}{8} \) is simply:
\[ \int_{0}^{\frac{\pi}{8}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{8}} = \frac{\pi}{8} - 0 = \frac{\pi}{8} \]

### Step 5: Combine the results
Now, substituting back into our separated integral:
\[ \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx = \frac{1}{2} - \frac{\pi}{8} \]

### Step 6: Simplify the result
To combine the terms, we can express \( \frac{1}{2} \) with a common denominator of 8:
\[ \frac{1}{2} = \frac{4}{8} \]
Thus:
\[ \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx = \frac{4}{8} - \frac{\pi}{8} = \frac{4 - \pi}{8} \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{8}} \tan^2(2x) \, dx = \frac{4 - \pi}{8} \]

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