`int_(pi)^(5pi//4) (sin 2x)/(cos^(4) x +sin^(4)x) dx`=

Options

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

Answer

To solve the integral

\[ I = \int_{\pi}^{\frac{5\pi}{4}} \frac{\sin 2x}{\cos^4 x + \sin^4 x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Denominator
We can rewrite the denominator \(\cos^4 x + \sin^4 x\) using the identity:

\[ a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2. \]

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

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

Thus, we have:

\[ I = \int_{\pi}^{\frac{5\pi}{4}} \frac{\sin 2x}{1 - 2\cos^2 x \sin^2 x} \, dx. \]

### Step 2: Use the Identity for \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). Therefore, we can rewrite the integral as:

\[ I = \int_{\pi}^{\frac{5\pi}{4}} \frac{2 \sin x \cos x}{1 - 2\cos^2 x \sin^2 x} \, dx. \]

### Step 3: Substitution
Let \(u = \sin 2x\). Then, the derivative \(du = 2 \cos 2x \, dx\), or \(dx = \frac{du}{2 \cos 2x}\).

Now, we need to change the limits of integration. When \(x = \pi\), \(u = \sin 2\pi = 0\), and when \(x = \frac{5\pi}{4}\), \(u = \sin \frac{5\pi}{2} = 1\).

### Step 4: Change of Variables
Now substituting into the integral gives:

\[ I = \int_{0}^{1} \frac{u}{1 - 2\left(\frac{1-u^2}{2}\right)} \cdot \frac{du}{2 \sqrt{1-u^2}}. \]

### Step 5: Simplify the Integral
The denominator simplifies to:

\[ 1 - (1 - u^2) = u^2. \]

Thus, the integral becomes:

\[ I = \int_{0}^{1} \frac{u}{u^2} \cdot \frac{du}{2 \sqrt{1-u^2}} = \frac{1}{2} \int_{0}^{1} \frac{1}{\sqrt{1-u^2}} \, du. \]

### Step 6: Evaluate the Integral
The integral \(\int \frac{1}{\sqrt{1-u^2}} \, du\) is a standard integral that evaluates to \(\sin^{-1}(u)\):

\[ I = \frac{1}{2} \left[ \sin^{-1}(u) \right]_{0}^{1} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}. \]

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

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