Evaluate `intcos^(3)xdx.`

Answer

To evaluate the integral \( \int \cos^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \cos^3 x \, dx \]
We can express \( \cos^3 x \) as \( \cos^2 x \cdot \cos x \):
\[ I = \int \cos^2 x \cdot \cos x \, dx \]

### Step 2: Use the Pythagorean Identity
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the integral:
\[ I = \int (1 - \sin^2 x) \cos x \, dx \]

### Step 3: Substitute \( \sin x \)
Let \( t = \sin x \). Then, the derivative \( dt = \cos x \, dx \). This substitution gives us:
\[ I = \int (1 - t^2) \, dt \]

### Step 4: Separate the Integral
Now we can separate the integral:
\[ I = \int 1 \, dt - \int t^2 \, dt \]

### Step 5: Integrate Each Term
Now we integrate each term:
1. The integral of \( 1 \) is \( t \).
2. The integral of \( t^2 \) is \( \frac{t^3}{3} \).

Thus, we have:
\[ I = t - \frac{t^3}{3} + C \]

### Step 6: Substitute Back \( t = \sin x \)
Now we substitute back \( t = \sin x \):
\[ I = \sin x - \frac{\sin^3 x}{3} + C \]

### Final Answer
The final result for the integral is:
\[ \int \cos^3 x \, dx = \sin x - \frac{\sin^3 x}{3} + C \]