\int \:sin^4\left(x\right)\:cos^2\:\left(x\right)

To evaluate the integral \( \int \sin^4(x) \cos^2(x) \, dx \), we can simplify the integrand using trigonometric identities.

### Step 1: Use power-reducing identities
We start by converting powers of sine and cosine using the following identities:

\[
\sin^2(x) = \frac{1 - \cos(2x)}{2}
\quad \text{and} \quad
\cos^2(x) = \frac{1 + \cos(2x)}{2}.
\]

Thus, we rewrite \( \sin^4(x) \cos^2(x) \) as:

\[
\sin^4(x) = (\sin^2(x))^2 = \left( \frac{1 - \cos(2x)}{2} \right)^2,
\]
so the integrand becomes:
\[
\int \left( \frac{1 - \cos(2x)}{2} \right)^2 \cdot \frac{1 + \cos(2x)}{2} \, dx.
\]

### Step 2: Expand the product
First, expand the square \( \left( \frac{1 - \cos(2x)}{2} \right)^2 \):

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

Now multiply this by \( \frac{1 + \cos(2x)}{2} \):

\[
\frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \cdot \frac{1 + \cos(2x)}{2} = \frac{1}{8} \left( (1 - 2\cos(2x) + \cos^2(2x))(1 + \cos(2x)) \right).
\]

### Step 3: Expand further
Now expand the expression inside the parentheses:

\[
(1 - 2\cos(2x) + \cos^2(2x))(1 + \cos(2x)) = 1 + \cos(2x) - 2\cos(2x) - 2\cos^2(2x) + \cos^2(2x) + \cos^3(2x).
\]

Simplify this:

\[
1 - \cos(2x) - \cos^2(2x) + \cos^3(2x).
\]

### Step 4: Split the integral
Now the integrand is:

\[
\frac{1}{8} \int \left( 1 - \cos(2x) - \cos^2(2x) + \cos^3(2x) \right) dx.
\]

We can split this into four simpler integrals:

\[
\frac{1}{8} \left( \int 1 \, dx - \int \cos(2x) \, dx - \int \cos^2(2x) \, dx + \int \cos^3(2x) \, dx \right).
\]

### Step 5: Evaluate each integral
1. \( \int 1 \, dx = x \).
2. \( \int \cos(2x) \, dx = \frac{\sin(2x)}{2} \).
3. \( \int \cos^2(2x) \, dx \) uses the power-reducing identity \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \), so:
   \[
   \int \cos^2(2x) \, dx = \frac{1}{2} \int 1 + \cos(4x) \, dx = \frac{x}{2} + \frac{\sin(4x)}{8}.
   \]
4. \( \int \cos^3(2x) \, dx \) uses the identity \( \cos^3(2x) = \cos(2x)(\frac{1 + \cos(4x)}{2}) \), which can be expanded and integrated term by term. However, this step involves a bit more algebra, and the result is:

\[
\int \cos^3(2x) \, dx = \frac{3 \sin(2x)}{8} + \frac{\sin(6x)}{48}.
\]

### Final Answer:
The full solution is:

\[
\frac{1}{8} \left( x - \frac{\sin(2x)}{2} - \left( \frac{x}{2} + \frac{\sin(4x)}{8} \right) + \left( \frac{3\sin(2x)}{8} + \frac{\sin(6x)}{48} \right) \right) + C.
\]

This can be simplified further, but this is the general structure.

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you apply trigonometric identities to simplify integrals?
2. What is the power-reducing identity for higher powers of sine and cosine?
3. How do you handle integrals involving products of trigonometric functions?
4. What are some strategies for solving integrals of powers of trigonometric functions?
5. How do you simplify the integration of cubic trigonometric functions?

**Tip:** Using symmetry in trigonometric functions can simplify integrals, especially when the powers are even.

This step-by-step guide explains how to solve the integral \( \int \sin^4(x) \cos^2(x) \, dx \) by applying trigonometric identities and power-reducing formulas. Key techniques include expanding trigonometric powers and splitting the integral into simpler terms. Suitable for students studying advanced high school calculus or undergraduate mathematics.

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