The value of `int_(0)^(pi//2) (dx)/(1+tan^(3) x)` is

Options

A. 0
B. 1
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^3 x} \), we can use a symmetry property of definite integrals. Here are the steps:

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^3 x} \]

### Step 2: Use the substitution \( x = \frac{\pi}{2} - t \)
Using the substitution \( x = \frac{\pi}{2} - t \), we have \( dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = \frac{\pi}{2} \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)

Thus, the integral becomes:
\[ I = \int_{\frac{\pi}{2}}^{0} \frac{-dt}{1 + \tan^3\left(\frac{\pi}{2} - t\right)} \]
Using the identity \( \tan\left(\frac{\pi}{2} - t\right) = \cot t \), we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dt}{1 + \cot^3 t} \]

### Step 3: Rewrite \( \cot^3 t \)
Recall that \( \cot t = \frac{1}{\tan t} \), so:
\[ \cot^3 t = \frac{1}{\tan^3 t} \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dt}{1 + \frac{1}{\tan^3 t}} = \int_{0}^{\frac{\pi}{2}} \frac{\tan^3 t}{\tan^3 t + 1} dt \]

### Step 4: Combine the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^3 x} \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\tan^3 x}{\tan^3 x + 1} dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{\tan^3 x + 1} \right) dx \]

### Step 5: Simplify the integrand
The integrand simplifies as follows:
\[ \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{\tan^3 x + 1} = \frac{1 + \tan^3 x}{1 + \tan^3 x} = 1 \]
Thus, we have:
\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \]

### Step 6: Solve for \( I \)
Dividing both sides by 2 gives:
\[ I = \frac{\pi}{4} \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan^3 x} = \frac{\pi}{4} \]