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The integral you provided is:

\[
\int \left( 1 - \frac{1}{y} \right) \cos(y - \ln y) \, dy
\]

Let's go through the process of solving this integral step by step.

### Step 1: Distribute the terms
First, distribute the term \( \left( 1 - \frac{1}{y} \right) \) to get two integrals:

\[
\int \left( 1 - \frac{1}{y} \right) \cos(y - \ln y) \, dy = \int \cos(y - \ln y) \, dy - \int \frac{1}{y} \cos(y - \ln y) \, dy
\]

### Step 2: Substitution
In both integrals, the expression \( y - \ln y \) suggests using substitution. Let's make the substitution:

\[
u = y - \ln y
\]
Taking the derivative of both sides with respect to \( y \):

\[
du = \left( 1 - \frac{1}{y} \right) dy
\]

This substitution simplifies both integrals significantly.

Now the integral becomes:

\[
\int \cos(u) \, du
\]

### Step 3: Integration
The integral of \( \cos(u) \) is straightforward:

\[
\int \cos(u) \, du = \sin(u) + C
\]

### Step 4: Back-substitute \( u = y - \ln y \)
Finally, substitute back for \( u \):

\[
\sin(y - \ln y) + C
\]

Thus, the solution to the integral is:

\[
\sin(y - \ln y) + C
\]

If you need further clarifications or have additional questions, feel free to ask!

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Here are 5 related questions to deepen your understanding:

1. How can you determine when substitution is a useful method for integration?
2. What is the derivative of \( \ln y \), and why is it important for solving this integral?
3. How does the chain rule apply when using substitution in integrals?
4. What are some other common trigonometric integrals?
5. Could this integral be evaluated using integration by parts? Why or why not?

**Tip:** When performing integration, it's often useful to check if a substitution simplifies the integrand, especially when you see expressions involving sums or differences like \( y - \ln y \).

Discover how to solve the integral ∫(1 - 1/y) cos(y - ln y) dy using the substitution method. Learn the step-by-step process of simplifying the integral to ∫cos(u) du and finding the final result, sin(y - ln y) + C. Suitable for undergraduate calculus students.

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