Evaluate: `int(dx)/(sin^4x+sin^2xcos^2x+cos^4x)`

Answer

To evaluate the integral

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

we will follow these steps:

### Step 1: Rewrite the Denominator
We start by rewriting the denominator:

\[ \sin^4 x + \cos^4 x + \sin^2 x \cos^2 x. \]

Using the identity \(a^2 + b^2 = (a + b)^2 - 2ab\), we can express \(\sin^4 x + \cos^4 x\) as:

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

Thus, the denominator becomes:

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

### Step 2: Substitute with Trigonometric Identities
We can use the identity \(\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)\):

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

Thus, our integral becomes:

\[ I = \int \frac{dx}{1 - \frac{1}{4} \sin^2(2x)}. \]

### Step 3: Simplify the Integral
We can rewrite this integral using the formula for the integral of the form \(\int \frac{dx}{a - b \sin^2(kx)}\):

\[ I = \int \frac{dx}{\frac{4}{4} - \frac{1}{4} \sin^2(2x)} = 4 \int \frac{dx}{4 - \sin^2(2x)}. \]

### Step 4: Use the Substitution
Let \(u = 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\):

\[ I = 4 \cdot \frac{1}{2} \int \frac{du}{4 - \sin^2(u)} = 2 \int \frac{du}{4 - \sin^2(u)}. \]

### Step 5: Further Simplify the Integral
Using the identity \(4 - \sin^2(u) = 4 \cos^2(u) + 3\):

\[ I = 2 \int \frac{du}{4 - \sin^2(u)} = 2 \int \frac{du}{4 - \sin^2(u)}. \]

### Step 6: Use the Integral Formula
The integral can be evaluated using the formula:

\[ \int \frac{dx}{a^2 - b^2 \sin^2(x)} = \frac{1}{a} \tan^{-1}\left(\frac{b \sin(x)}{a}\right) + C. \]

Here, \(a = 2\) and \(b = 1\):

\[ I = 2 \cdot \frac{1}{2} \tan^{-1}\left(\frac{\sin(u)}{2}\right) + C = \tan^{-1}\left(\frac{\sin(2x)}{2}\right) + C. \]

### Final Step: Substitute Back
Substituting back for \(u\):

\[ I = \tan^{-1}\left(\frac{\sin(2x)}{2}\right) + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = \tan^{-1}\left(\frac{\sin(2x)}{2}\right) + C. \]