We are tasked with evaluating the integral:

$\int \frac{\cos(5x+1)}{\sin(5x+1)+6} \, dx$

Solution:

Let us solve step by step:

1. Substitution:
   Let: $u = \sin(5x+1) + 6$ Then: $du = 5\cos(5x+1) \, dx$ Therefore: $\cos(5x+1) \, dx = \frac{1}{5} \, du$

2. Rewriting the integral:
   Substitute $u$ and $du$ into the integral: $\int \frac{\cos(5x+1)}{\sin(5x+1)+6} \, dx = \int \frac{1}{u} \cdot \frac{1}{5} \, du$

   Simplify: $\int \frac{\cos(5x+1)}{\sin(5x+1)+6} \, dx = \frac{1}{5} \int \frac{1}{u} \, du$

3. Evaluate the integral:
   The integral of $\frac{1}{u}$ is: $\int \frac{1}{u} \, du = \ln|u|$

   So: $\frac{1}{5} \int \frac{1}{u} \, du = \frac{1}{5} \ln|u| + C$

4. Substitute back $u = \sin(5x+1) + 6$:
   Replace $u$ with the original substitution: $\int \frac{\cos(5x+1)}{\sin(5x+1)+6} \, dx = \frac{1}{5} \ln|\sin(5x+1) + 6| + C$

Final Answer:

$\boxed{\int \frac{\cos(5x+1)}{\sin(5x+1)+6} \, dx = \frac{1}{5} \ln|\sin(5x+1) + 6| + C}$

Would you like further clarification on any part of the solution?

5 Related Questions:

1. How can substitution be used effectively in other types of integrals?
2. What is the general formula for integrating $\frac{\cos(ax+b)}{\sin(ax+b)+c}$?
3. How does the constant factor (like $5$) in derivatives affect substitution steps?
4. What is the role of natural logarithms in integrals of the form $\int \frac{1}{u} \, du$?
5. Can we evaluate this integral using a different technique, such as trigonometric identities?

Tip:

Always check your substitutions carefully and ensure that the differential $du$ matches the integrand for smooth solving!