Answer is: option1

\(\frac{1}{n} f(x) \sin(nx) - \frac{1}{n} \int f'(x) \sin(nx) \, dx\)

Solution:

We are tasked with solving the integral:

\[ \int f(x) \cos(nx) \, dx \]

Step 1: Choose \( u \) and \( dv \)

We choose:

\[ u = f(x) \quad \Rightarrow \quad du = f'(x) \, dx \]

\[ dv = \cos(nx) \, dx \quad \Rightarrow \quad v = \frac{1}{n} \sin(nx) \]

Step 2: Apply the Integration by Parts Formula

The integration by parts formula is:

\[ \int u \, dv = u v - \int v \, du \]

Substitute \( u = f(x) \), \( du = f'(x) \, dx \), and \( v = \frac{1}{n} \sin(nx) \):

\[ \int f(x) \cos(nx) \, dx = f(x) \cdot \frac{1}{n} \sin(nx) - \int \frac{1}{n} \sin(nx) \cdot f'(x) \, dx \]

Step 3: Simplify the Expression

This simplifies to:

\[ \int f(x) \cos(nx) \, dx = \frac{1}{n} f(x) \sin(nx) - \frac{1}{n} \int f'(x) \sin(nx) \, dx \]

Final Answer

The final solution is:

\[ \int f(x) \cos(nx) \, dx = \frac{1}{n} f(x) \sin(nx) - \frac{1}{n} \int f'(x) \sin(nx) \, dx \]