`int(sin4x)/(1+sin^(4)2x)dx=`

Options

A. `-(1)/(2)tan^(-1)(sin^(2)2x)+c`
B. `(1)/(2)tan^(-1)(sin^(2)2x)+c`
C. `tan^(-1)(sin^(2)2x)+c`
D. `cot^(-1)(sin^(2)2x)+c`

Answer

To solve the integral \( \int \frac{\sin 4x}{1 + \sin^4 2x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \sin 4x \) using the double angle formula:
\[ \sin 4x = 2 \sin 2x \cos 2x \]
Thus, we can rewrite the integral as:
\[ \int \frac{2 \sin 2x \cos 2x}{1 + \sin^4 2x} \, dx \]

### Step 2: Substitute \( t = \sin^2 2x \)
Let us make the substitution \( t = \sin^2 2x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = 2 \sin 2x \cos 2x \cdot 2 = 4 \sin 2x \cos 2x \]
This implies:
\[ dt = 4 \sin 2x \cos 2x \, dx \quad \Rightarrow \quad \sin 2x \cos 2x \, dx = \frac{dt}{4} \]

### Step 3: Substitute in the integral
Now we can substitute \( \sin 2x \cos 2x \, dx \) in our integral:
\[ \int \frac{2 \sin 2x \cos 2x}{1 + \sin^4 2x} \, dx = \int \frac{2 \cdot \frac{dt}{4}}{1 + t^2} = \int \frac{dt}{2(1 + t^2)} \]

### Step 4: Integrate
The integral \( \int \frac{dt}{1 + t^2} \) is a standard integral that results in:
\[ \int \frac{dt}{1 + t^2} = \tan^{-1}(t) + C \]
Thus, we have:
\[ \int \frac{dt}{2(1 + t^2)} = \frac{1}{2} \tan^{-1}(t) + C \]

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = \sin^2 2x \):
\[ \frac{1}{2} \tan^{-1}(\sin^2 2x) + C \]

### Final Result
Therefore, the final result of the integral is:
\[ \int \frac{\sin 4x}{1 + \sin^4 2x} \, dx = \frac{1}{2} \tan^{-1}(\sin^2 2x) + C \]

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