The integral provided is:

$\int \sin(x) \sec^3(x) \, dx$

Solution:

To solve this, we use trigonometric identities and substitution. The steps are:

1. Rewriting secant using its identity: Recall that $\sec(x) = \frac{1}{\cos(x)}$. Thus: $\sec^3(x) = \frac{1}{\cos^3(x)}.$ Substituting this, the integral becomes: $\int \sin(x) \cdot \frac{1}{\cos^3(x)} \, dx = \int \frac{\sin(x)}{\cos^3(x)} \, dx.$

2. Substitution: Let $u = \cos(x)$. Then $du = -\sin(x) \, dx$, or $-du = \sin(x) \, dx$. Substitute into the integral: $\int \frac{\sin(x)}{\cos^3(x)} \, dx = \int \frac{-1}{u^3} \, du.$

3. Simplify and integrate: The integral $\int \frac{-1}{u^3} \, du$ can be rewritten as: $-\int u^{-3} \, du.$ The formula for integrating $u^n$ is: $\int u^n \, du = \frac{u^{n+1}}{n+1}, \quad \text{for } n \neq -1.$ Applying this: $-\int u^{-3} \, du = -\frac{u^{-3+1}}{-3+1} = -\frac{u^{-2}}{-2} = \frac{u^{-2}}{2}.$

4. Back-substitute $u = \cos(x)$: Recall that $u = \cos(x)$. Thus: $\frac{u^{-2}}{2} = \frac{1}{2u^2} = \frac{1}{2\cos^2(x)} = \frac{\sec^2(x)}{2}.$

5. Add the constant of integration: The final answer is: $\int \sin(x) \sec^3(x) \, dx = \frac{\sec^2(x)}{2} + C,$ where $C$ is the constant of integration.

Do you have questions or need further clarification?

Here are five related questions to expand your understanding:

1. How do we derive the formula for integrating powers of $u$?
2. Why is substitution useful for trigonometric integrals?
3. Can you integrate $\int \cos(x) \sec^3(x) \, dx$ for comparison?
4. How does this integral change if a different substitution is used?
5. What are some other trigonometric integral techniques?

Tip: Always check if a substitution simplifies the integral before using more advanced methods!