Evaluate :
`int sin^(2)xcos^(4)xdx`

Answer

To evaluate the integral \( \int \sin^2 x \cos^4 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We can express \( \cos^4 x \) as \( \cos^2 x \cdot \cos^2 x \):
\[ \int \sin^2 x \cos^4 x \, dx = \int \sin^2 x \cos^2 x \cos^2 x \, dx \]

### Step 2: Use the identity for \( \sin^2 x \cos^2 x \)
We know that:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite the integral as:
\[ \int \sin^2 x \cos^4 x \, dx = \int \frac{1}{4} \sin^2(2x) \cos^2 x \, dx \]

### Step 3: Introduce a factor of \( \frac{1}{8} \)
To facilitate integration, we multiply and divide by 8:
\[ = \frac{1}{8} \int 8 \sin^2 x \cos^2 x \cos^2 x \, dx \]

### Step 4: Substitute \( u = 2x \)
Let \( u = 2x \), then \( du = 2dx \) or \( dx = \frac{du}{2} \):
\[ = \frac{1}{8} \int 4 \sin^2 x \cos^2 x \cdot \cos^2 x \cdot \frac{du}{2} \]

### Step 5: Simplify the integral
Now we can use the double angle identity for sine:
\[ = \frac{1}{16} \int 4 \sin^2 x \cos^2 x \, \cos^2 x \, du \]

### Step 6: Expand the integral
Using \( \cos^2 x = \frac{1 + \cos(2x)}{2} \):
\[ = \frac{1}{16} \int (1 - \cos(4x))(1 + \cos(2x)) \, dx \]

### Step 7: Distribute and integrate
Distributing gives:
\[ = \frac{1}{16} \left( \int (1 - \cos(4x) + \cos(2x) - \frac{1}{2} \cos(6x)) \, dx \right) \]

### Step 8: Integrate each term separately
The integral of \( 1 \) is \( x \), the integral of \( \cos(kx) \) is \( \frac{\sin(kx)}{k} \):
\[ = \frac{1}{16} \left( x - \frac{1}{4} \sin(4x) + \frac{1}{2} \sin(2x) - \frac{1}{12} \sin(6x) + C \right) \]

### Step 9: Combine and simplify
Now we can combine the results:
\[ = \frac{1}{16} x - \frac{1}{64} \sin(4x) + \frac{1}{32} \sin(2x) - \frac{1}{192} \sin(6x) + C \]

### Final Answer
Thus, the final answer is:
\[ \frac{1}{16} x - \frac{1}{64} \sin(4x) + \frac{1}{32} \sin(2x) - \frac{1}{192} \sin(6x) + C \]