Let's break down the steps in this problem based on the provided image.

Step 1:

The integral to evaluate is:

$I = \int \text{sech}(u) \, du = \int \frac{1}{\cosh(u)} \, du$

We are instructed to use the hyperbolic substitution:

$t = \sinh(u)$

Step 2:

To perform this substitution, we need to find $\frac{dt}{du}$.

Given that $t = \sinh(u)$, the derivative of $\sinh(u)$ is:

$\frac{dt}{du} = \cosh(u)$

Step 3:

The integral $I$ can now be rewritten in terms of $t$. We substitute $\cosh(u) \, du = dt$ into the original integral:

$I = \int \frac{1}{\cosh(u)} \, du = \int \frac{1}{\cosh(u)} \cdot \cosh(u) \, dt = \int \frac{1}{t} \, dt$

Step 4:

The integral becomes:

$I = \int \frac{1}{t} \, dt$

This is a standard integral, which evaluates to:

$I = \ln|t| + C$

Step 5:

Since $t = \sinh(u)$, we substitute back to express the result in terms of $u$:

$I = \ln|\sinh(u)| + C$

Thus, the final result is:

$I = \ln|\sinh(u)| + C$

Would you like more details or have any questions about this solution?

Here are 5 related questions to deepen your understanding:

1. What are the properties of the hyperbolic functions $\sinh(u)$ and $\cosh(u)$?
2. How would you evaluate $\int \text{csch}(u) \, du$?
3. What is the difference between trigonometric and hyperbolic substitutions in integrals?
4. Can the integral $\int \cosh(u) \, du$ be solved using a similar substitution technique?
5. How does the substitution $t = \sinh(u)$ simplify integrals involving $\cosh(u)$?

Tip: Understanding hyperbolic identities, like $\cosh^2(u) - \sinh^2(u) = 1$, can make integrals involving hyperbolic functions easier to solve!