Evaluate : `int_(0)^((pi)/(2)) (x sin x.cosx)/(sin^(4)x+cos^(4)x)dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx, \]

we can use the property of definite integrals that states:

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

### Step 1: Apply the property

Using the property, we can rewrite the integral \(I\):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x) \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)}{\sin^4(\frac{\pi}{2} - x) + \cos^4(\frac{\pi}{2} - x)} \, dx. \]

### Step 2: Simplify the expression

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

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - x\right) \cos x \sin x}{\cos^4 x + \sin^4 x} \, dx. \]

### Step 3: Combine the two integrals

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

1. \(I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx\)
2. \(I = \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - x\right) \cos x \sin x}{\cos^4 x + \sin^4 x} \, dx\)

Adding these two equations:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}\right) \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 4: Factor out constants

We can factor out \(\frac{\pi}{2}\):

\[ 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 5: Evaluate the integral

Now, we need to evaluate the integral:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 6: Simplify the denominator

Notice that:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x. \]

Thus, the integral becomes:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1 - 2\sin^2 x \cos^2 x} \, dx. \]

### Step 7: Use substitution

Let \(u = \sin^2 x\), then \(du = 2\sin x \cos x \, dx\). The limits change from \(0\) to \(1\):

\[ \int_{0}^{1} \frac{1}{1 - 2u(1-u)} \frac{du}{2} = \frac{1}{2} \int_{0}^{1} \frac{1}{1 - 2u + 2u^2} \, du. \]

### Step 8: Evaluate the integral

This can be evaluated using partial fractions or recognizing it as a standard integral.

### Step 9: Final calculation

After evaluating the integral, we find:

\[ I = \frac{\pi^2}{16}. \]

### Conclusion

Thus, the final result is:

\[ \boxed{\frac{\pi^2}{16}}. \]