`int_(0)^(pi//4)(sin 2 x)/(sin^(4) x + cos^(4) x ) dx = `

Options

A. `pi/2`
B. `pi/4`
C. `(3pi)/2`
D. `(3pi)/4`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the Denominator
We start by simplifying the denominator \(\sin^4 x + \cos^4 x\). We can use the identity:

\[ a^2 + b^2 = (a + b)^2 - 2ab. \]

Let \(a = \sin^2 x\) and \(b = \cos^2 x\). Then we have:

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

Since \(\sin^2 x + \cos^2 x = 1\), we get:

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

Using the identity \(\sin 2x = 2\sin x \cos x\), we can express \(\sin^2 x \cos^2 x\) as:

\[ \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}. \]

Thus, we can rewrite the denominator:

\[ \sin^4 x + \cos^4 x = 1 - \frac{\sin^2 2x}{2}. \]

### Step 2: Rewrite the Integral
Now substituting this back into the integral, we have:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2x}{1 - \frac{\sin^2 2x}{2}} \, dx. \]

This can be simplified to:

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

### Step 3: Substitution
Let \(t = \cos 2x\). Then, we have:

\[ dt = -2\sin 2x \, dx \implies dx = \frac{-dt}{2\sin 2x}. \]

When \(x = 0\), \(t = \cos 0 = 1\) and when \(x = \frac{\pi}{4}\), \(t = \cos \frac{\pi}{2} = 0\). Thus, the limits change from \(1\) to \(0\).

Substituting into the integral gives:

\[ I = \int_{1}^{0} \frac{2\sin 2x}{2 - \sin^2 2x} \cdot \frac{-dt}{2\sin 2x} = \int_{0}^{1} \frac{dt}{2 - \sin^2 2x}. \]

### Step 4: Express \(\sin^2 2x\) in terms of \(t\)
Since \(t = \cos 2x\), we have \(\sin^2 2x = 1 - t^2\). Therefore, the integral becomes:

\[ I = \int_{0}^{1} \frac{dt}{2 - (1 - t^2)} = \int_{0}^{1} \frac{dt}{1 + t^2}. \]

### Step 5: Evaluate the Integral
The integral \(\int \frac{dt}{1 + t^2}\) is known to be:

\[ \int \frac{dt}{1 + t^2} = \tan^{-1}(t) + C. \]

Thus,

\[ I = \left[ \tan^{-1}(t) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Answer
Therefore, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]