Find the value of `int_(0)^(2pi)1/(1+tan^(4)x)dx`

Explanation

To solve the integral \( I = \int_{0}^{2\pi} \frac{1}{1 + \tan^4 x} \, dx \), we will follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{2\pi} \frac{1}{1 + \tan^4 x} \, dx \]

### Step 2: Use the Symmetry Property of Integrals
We can use the property of definite integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In this case, we can express \( I \) as:
\[ I = \int_{0}^{\pi} \frac{1}{1 + \tan^4 x} \, dx + \int_{\pi}^{2\pi} \frac{1}{1 + \tan^4 x} \, dx \]
Using the substitution \( x = 2\pi - t \) in the second integral, we get:
\[ \int_{\pi}^{2\pi} \frac{1}{1 + \tan^4 x} \, dx = \int_{0}^{\pi} \frac{1}{1 + \tan^4(2\pi - t)} \, dt \]
Since \( \tan(2\pi - t) = -\tan(t) \), we have:
\[ \tan^4(2\pi - t) = \tan^4(t) \]
Thus,
\[ I = \int_{0}^{\pi} \frac{1}{1 + \tan^4 x} \, dx + \int_{0}^{\pi} \frac{1}{1 + \tan^4 x} \, dx = 2 \int_{0}^{\pi} \frac{1}{1 + \tan^4 x} \, dx \]
So,
\[ I = 2 \int_{0}^{\pi} \frac{1}{1 + \tan^4 x} \, dx \]

### Step 3: Change of Variables
Next, we will change the variable in the integral:
\[ I = 2 \int_{0}^{\pi} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx \]
This is achieved by rewriting \( \tan^4 x \) in terms of sine and cosine.

### Step 4: Use the Symmetry Again
Now, we can express the integral as:
\[ I = 2 \int_{0}^{\pi} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx \]
This simplifies to:
\[ I = 2 \int_{0}^{\pi} 1 \, dx = 2 \times \pi = 2\pi \]

### Step 5: Final Calculation
Now we have:
\[ I = 2 \int_{0}^{\pi} 1 \, dx = 2 \cdot \left( \frac{\pi}{2} \right) = \pi \]

### Conclusion
Thus, the value of the integral is:
\[ \boxed{\pi} \]