Evaluate : `int_(0)^(pi/2)(sin^2x-cos^2x)/(sin^3x+cos^3x)dx`

Answer

To evaluate the integral

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

we can use a property of definite integrals. This property states that

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

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

Here, we have \( a = 0 \) and \( b = \frac{\pi}{2} \). Thus, we can rewrite the integral as:

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

### Step 2: Simplify the expressions

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

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

### Step 3: Combine the two integrals

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

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

Adding these two equations:

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

### Step 4: Simplify the numerator

Notice that the numerator simplifies to zero:

\[ \sin^2 x - \cos^2 x + \cos^2 x - \sin^2 x = 0. \]

Thus, we have:

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

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

Since \( 2I = 0 \), we conclude that:

\[ I = 0. \]

### Final Answer

The value of the integral is

\[ \boxed{0}. \]