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 find the value of 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 a property of definite integrals. This property states that

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

In our case, we will let \( a = \frac{\pi}{2} \) and replace \( x \) with \( \frac{\pi}{2} - x \).

### Step 1: Apply the property of definite integrals

Using the property, we have:

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

### Step 2: Simplify the trigonometric functions

Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we can rewrite the integral:

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

### Step 3: Combine the two expressions for \( I \)

Now we have two expressions for \( I \):

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

### Step 4: Add the two equations

Adding both expressions for \( I \):

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

The terms \( \sin^3 x \) and \( -\sin^3 x \), as well as \( \cos^3 x \) and \( -\cos^3 x \), cancel each other out:

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

### Step 5: Solve for \( I \)

Thus, we find:

\[ 2I = 0 \implies I = 0. \]

### Final Answer

The value of the integral is

\[ \boxed{0}. \]