The value of the integral `I=int_(0)^((pi)/(2))(sin^(3)x -cos^(3)x)/(1+sin^(6)x cos^(6)x)dx` is equal to

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x - \cos^3 x}{1 + \sin^6 x \cos^6 x} \, dx, \]

we can use the property of definite integrals. Specifically, we can use the substitution \( x = \frac{\pi}{2} - t \).

### Step 1: Apply the substitution
Let \( x = \frac{\pi}{2} - t \). Then, \( dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = \frac{\pi}{2} \).
- When \( x = \frac{\pi}{2} \), \( t = 0 \).

Thus, we can rewrite the integral as:

\[ I = \int_{\frac{\pi}{2}}^{0} \frac{\sin^3\left(\frac{\pi}{2} - t\right) - \cos^3\left(\frac{\pi}{2} - t\right)}{1 + \sin^6\left(\frac{\pi}{2} - t\right) \cos^6\left(\frac{\pi}{2} - t\right)} (-dt). \]

### Step 2: Simplify the integrand
Using the identities \( \sin\left(\frac{\pi}{2} - t\right) = \cos t \) and \( \cos\left(\frac{\pi}{2} - t\right) = \sin t \), we get:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 t - \sin^3 t}{1 + \cos^6 t \sin^6 t} dt. \]

### Step 3: Combine the two integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x - \cos^3 x}{1 + \sin^6 x \cos^6 x} \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x - \sin^3 x}{1 + \cos^6 x \sin^6 x} \, dx \)

Adding these two equations, we have:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin^3 x - \cos^3 x) + (\cos^3 x - \sin^3 x)}{1 + \sin^6 x \cos^6 x} \, dx. \]

### Step 4: Simplify the expression
Notice that the terms \( \sin^3 x - \cos^3 x \) and \( \cos^3 x - \sin^3 x \) cancel each other out:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{0}{1 + \sin^6 x \cos^6 x} \, dx = 0. \]

### Step 5: Conclude the value of the integral
Thus, we find that

\[ I = 0. \]

### Final Answer:
The value of the integral \( I \) is

\[ \boxed{0}. \]