`int_(0)^(pi) (dx)/(1 + tan^(4)x)`=

Options

A. 0
B. `pi//6`
C. `pi//4`
D. `pi//2`

Answer

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

### Step 1: Use the property of definite integrals
We will use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we have \( a = \pi \). Thus, we can write:
\[ I = \int_{0}^{\pi} \frac{dx}{1 + \tan^4 x} = \int_{0}^{\pi} \frac{dx}{1 + \tan^4(\pi - x)} \]
Since \( \tan(\pi - x) = -\tan(x) \), we have:
\[ \tan^4(\pi - x) = \tan^4(x) \]
So, the integral remains the same:
\[ I = \int_{0}^{\pi} \frac{dx}{1 + \tan^4 x} \]

### Step 2: Split the integral
Now, we can express \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^4 x} + \int_{\frac{\pi}{2}}^{\pi} \frac{dx}{1 + \tan^4 x} \]
Using the substitution \( x = \pi - u \) in the second integral, we have:
\[ dx = -du \quad \text{and when } x = \frac{\pi}{2}, u = \frac{\pi}{2} \text{ and when } x = \pi, u = 0 \]
Thus, the second integral becomes:
\[ \int_{\frac{\pi}{2}}^{\pi} \frac{dx}{1 + \tan^4 x} = \int_{0}^{\frac{\pi}{2}} \frac{-du}{1 + \tan^4(\pi - u)} = \int_{0}^{\frac{\pi}{2}} \frac{du}{1 + \tan^4 u} \]
So, we can write:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^4 x} \]

### Step 3: Change of variables
Next, we can use the substitution \( \tan x = t \), which gives \( dx = \frac{dt}{1 + t^2} \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \) which corresponds to \( t = 0 \) to \( t = \infty \):
\[ I = 2 \int_{0}^{\infty} \frac{1}{1 + t^4} \cdot \frac{dt}{1 + t^2} \]

### Step 4: Simplifying the integral
Now, we can simplify the integral:
\[ I = 2 \int_{0}^{\infty} \frac{dt}{(1 + t^4)(1 + t^2)} \]
This integral can be evaluated using partial fractions or known integral results.

### Step 5: Final evaluation
After evaluating the integral, we find:
\[ I = \frac{\pi}{2} \]

Thus, the final result is:
\[ \int_{0}^{\pi} \frac{dx}{1 + \tan^4 x} = \frac{\pi}{2} \]