The value of `int_0^1 (x^(3))/(1+x^(8))dx` is

Options

A. `(pi)/(4)`
B. `(pi)/(8)`
C. `(pi)/(16)`
D. none of these

Answer

To solve the integral \( I = \int_0^1 \frac{x^3}{1+x^8} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Substitution
Let \( t = x^4 \). Then, differentiating both sides gives:
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]
Also, since \( t = x^4 \), we have \( x^3 = t^{3/4} \).

### Step 2: Change of Limits
When \( x = 0 \):
\[ t = 0^4 = 0 \]
When \( x = 1 \):
\[ t = 1^4 = 1 \]
Thus, the limits of integration remain from 0 to 1.

### Step 3: Rewrite the Integral
Now substituting \( x^3 \) and \( dx \) into the integral, we get:
\[ I = \int_0^1 \frac{t^{3/4}}{1 + (t^{1/4})^8} \cdot \frac{dt}{4t^{3/4}} = \int_0^1 \frac{1}{4(1 + t^2)} \, dt \]
The \( t^{3/4} \) terms cancel out.

### Step 4: Factor Out the Constant
We can factor out \( \frac{1}{4} \):
\[ I = \frac{1}{4} \int_0^1 \frac{1}{1 + t^2} \, dt \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{1}{1+t^2} \, dt \) is a standard integral that evaluates to \( \tan^{-1}(t) \):
\[ I = \frac{1}{4} \left[ \tan^{-1}(t) \right]_0^1 \]

### Step 6: Apply the Limits
Now we apply the limits:
\[ I = \frac{1}{4} \left( \tan^{-1}(1) - \tan^{-1}(0) \right) \]
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \tan^{-1}(0) = 0 \):
\[ I = \frac{1}{4} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{16} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{16}} \]