To solve the integral $\int \cos(2x - 7) \, \mathrm{d}x$, we can use the following steps:

1. Substitution: Let $u = 2x - 7$. Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = 2 \implies du = 2 \, dx$ This implies that $dx = \frac{du}{2}$.

2. Substitute: Replace $2x - 7$ with $u$ and $dx$ with $\frac{du}{2}$: $\int \cos(2x - 7) \, \mathrm{d}x = \int \cos(u) \cdot \frac{du}{2} = \frac{1}{2} \int \cos(u) \, du$

3. Integrate: The integral of $\cos(u)$ with respect to $u$ is $\sin(u)$: $\frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C$ where $C$ is the constant of integration.

4. Back-substitute: Replace $u$ with $2x - 7$ to get the final answer: $\frac{1}{2} \sin(2x - 7) + C$

So the final result is: $\int \cos(2x - 7) \, \mathrm{d}x = \frac{1}{2} \sin(2x - 7) + C$

Would you like any further details or explanations? Here are some related questions:

1. What is the integral of $\sin(2x - 7)$?
2. How do we integrate $\cos^2(2x - 7)$?
3. Can you explain the substitution method in more detail?
4. What happens if the coefficient of $x$ in the cosine function is not 2?
5. How would you differentiate $\sin(2x - 7)$?

Tip: When performing substitution in integrals, always remember to adjust the differential $dx$ accordingly to avoid mistakes.