` int _(0)^(pi//4) ( 4 sin 2 theta d theta )/(sin^(4) theta +cos^(4) theta )= `

Options

A. `pi/4`
B. `pi/2`
C. `pi`
D. None of these

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Rewrite \(\sin 2\theta\)
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{4 (2 \sin \theta \cos \theta) \, d\theta}{\sin^4 \theta + \cos^4 \theta} = \int_{0}^{\frac{\pi}{4}} \frac{8 \sin \theta \cos \theta \, d\theta}{\sin^4 \theta + \cos^4 \theta} \]

### Step 2: Simplify the Denominator
Next, we can factor the denominator:

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

Thus, we rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{8 \sin \theta \cos \theta \, d\theta}{1 - 2\sin^2 \theta \cos^2 \theta} \]

### Step 3: Use the Substitution
Let \(t = \tan \theta\), then \(d\theta = \frac{dt}{1+t^2}\). The limits change as follows:
- When \(\theta = 0\), \(t = 0\)
- When \(\theta = \frac{\pi}{4}\), \(t = 1\)

Also, we have:

\[ \sin \theta = \frac{t}{\sqrt{1+t^2}}, \quad \cos \theta = \frac{1}{\sqrt{1+t^2}} \]

Now substituting these into the integral:

\[ I = \int_{0}^{1} \frac{8 \cdot \frac{t}{\sqrt{1+t^2}} \cdot \frac{1}{\sqrt{1+t^2}} \cdot \frac{dt}{1+t^2}}{1 - 2\left(\frac{t^2}{1+t^2}\right)\left(\frac{1}{1+t^2}\right)} = \int_{0}^{1} \frac{8t \, dt}{(1+t^2)(1 - \frac{2t^2}{(1+t^2)^2})} \]

### Step 4: Simplify the Integral
The denominator simplifies to:

\[ 1 - \frac{2t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 2t^2}{(1+t^2)^2} = \frac{1 - t^2 + t^4}{(1+t^2)^2} \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{8t(1+t^2)^2 \, dt}{(1+t^2)(1 - t^2 + t^4)} = \int_{0}^{1} \frac{8t(1+t^2) \, dt}{1 - t^2 + t^4} \]

### Step 5: Evaluate the Integral
This integral can be evaluated using standard techniques or recognized as a form that leads to an arctangent function.

After evaluating, we find:

\[ I = 4 \tan^{-1}(1) - 4 \tan^{-1}(0) = 4 \cdot \frac{\pi}{4} - 0 = \pi \]

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

\[ \boxed{\pi} \]