The integral given is:

$\int_{3}^{7} \frac{\cosh(\ln(5t))}{t} \, dt$

Step-by-Step Solution:

1. Simplification of the Integrand:

We know that $\cosh(x)$ can be expressed as:

$\cosh(x) = \frac{e^x + e^{-x}}{2}$

Let's set $x = \ln(5t)$. So,

$\cosh(\ln(5t)) = \frac{e^{\ln(5t)} + e^{-\ln(5t)}}{2}$

Using the properties of logarithms and exponents, we have:

$e^{\ln(5t)} = 5t \quad \text{and} \quad e^{-\ln(5t)} = \frac{1}{5t}$

Thus,

$\cosh(\ln(5t)) = \frac{5t + \frac{1}{5t}}{2} = \frac{25t^2 + 1}{10t}$

2. Substitute into the Integral:

Now, substitute this expression back into the original integral:

$\int_{3}^{7} \frac{\frac{25t^2 + 1}{10t}}{t} \, dt = \frac{1}{10} \int_{3}^{7} \frac{25t^2 + 1}{t^2} \, dt$

This simplifies to:

$\frac{1}{10} \int_{3}^{7} \left( 25 + \frac{1}{t^2} \right) dt$

3. Evaluate the Integral:

The integral can now be broken up as:

$\frac{1}{10} \left( \int_{3}^{7} 25 \, dt + \int_{3}^{7} \frac{1}{t^2} \, dt \right)$

• The first integral is straightforward:

$\int_{3}^{7} 25 \, dt = 25[t]_{3}^{7} = 25(7 - 3) = 25 \times 4 = 100$

• The second integral is:

$\int_{3}^{7} \frac{1}{t^2} \, dt = \left[ -\frac{1}{t} \right]_{3}^{7} = -\frac{1}{7} + \frac{1}{3} = \frac{3 - 7}{21} = -\frac{4}{21}$

4. Final Calculation:

Now, substitute these values back into the expression:

$\frac{1}{10} \left( 100 - \frac{4}{21} \right) = \frac{1}{10} \left( 100 - \frac{4}{21} \right) = \frac{1}{10} \times \frac{2100 - 4}{21} = \frac{1}{10} \times \frac{2096}{21} = \frac{2096}{210}$

Simplifying this fraction:

$\frac{2096}{210} = 9.98 \quad (\text{approximately})$

Thus, the value of the integral is approximately $9.98$.

Do you need more details or have any questions?

Here are 5 related questions:

1. How do we simplify $\cosh(\ln(x))$ in general cases?
2. What are the properties of the hyperbolic cosine function ($\cosh$)?
3. Can we generalize the method for evaluating integrals with other hyperbolic functions?
4. How does the substitution technique help in integral evaluation?
5. How would the limits change the outcome if different values were used?

Tip: Always look for algebraic simplifications (like exponent properties) when dealing with logarithmic and hyperbolic functions in integrals.