Evaluate: `int(cos2xsin4x dx)/(cos^4x(1+cos^2 2x)`

Explanation

To evaluate the integral

\[ I = \int \frac{\cos 2x \sin 4x \, dx}{\cos^4 x (1 + \cos^2 2x)} \]

we will follow these steps:

### Step 1: Rewrite \(\sin 4x\)
We can use the identity \(\sin 4x = 2 \sin 2x \cos 2x\). Thus, we rewrite the integral:

\[ I = \int \frac{\cos 2x (2 \sin 2x \cos 2x) \, dx}{\cos^4 x (1 + \cos^2 2x)} = 2 \int \frac{\cos^2 2x \sin 2x \, dx}{\cos^4 x (1 + \cos^2 2x)} \]

### Step 2: Substitute \(\cos 2x\)
Let \(t = \cos 2x\). Then, we differentiate \(t\):

\[ dt = -2 \sin 2x \, dx \quad \Rightarrow \quad \sin 2x \, dx = -\frac{1}{2} dt \]

Substituting this into the integral gives:

\[ I = 2 \int \frac{t^2 \left(-\frac{1}{2} dt\right)}{\cos^4 x (1 + t^2)} = -\int \frac{t^2 \, dt}{\cos^4 x (1 + t^2)} \]

### Step 3: Rewrite \(\cos^4 x\)
Using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\), we have:

\[ \cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{(1 + t)^2}{4} \]

Thus, we can rewrite the integral as:

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

### Step 4: Partial Fraction Decomposition
We need to perform partial fraction decomposition on:

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

Assuming:

\[ \frac{t^2}{(1 + t)^2 (1 + t^2)} = \frac{A}{1 + t} + \frac{B}{(1 + t)^2} + \frac{Ct + D}{1 + t^2} \]

Multiplying through by the denominator \((1 + t)^2 (1 + t^2)\) and equating coefficients will allow us to solve for \(A\), \(B\), \(C\), and \(D\).

### Step 5: Integrate Each Term
Once we have the coefficients, we can integrate each term separately. The integrals will involve:

1. \(\int \frac{1}{1 + t} dt\)
2. \(\int \frac{1}{(1 + t)^2} dt\)
3. \(\int \frac{t}{1 + t^2} dt\)

### Step 6: Back Substitute
After integrating, we will substitute back \(t = \cos 2x\) to express the final result in terms of \(x\).

### Final Result
The final expression will be:

\[ I = -4 \left( \text{integrated terms} \right) + C \]

where \(C\) is the constant of integration.