We are tasked with evaluating the integral:

$\int \frac{\sin(\sqrt{x})}{\sqrt{x} \cdot \cos^3(\sqrt{x})} \, dx$

Step 1: Substitution

We begin by making the substitution $u = \sqrt{x}$, which implies that: $x = u^2 \quad \text{and} \quad dx = 2u \, du$ Also, note that $\sqrt{x} = u$, so the integral becomes:

$\int \frac{\sin(u)}{u \cdot \cos^3(u)} \cdot 2u \, du$

This simplifies to:

$2 \int \frac{\sin(u)}{\cos^3(u)} \, du$

Step 2: Use Trigonometric Identity

We now focus on the integral: $\int \frac{\sin(u)}{\cos^3(u)} \, du$ We recognize that this can be solved by using a simple substitution. Since the derivative of $\cos(u)$ is $-\sin(u)$, let’s make the substitution:

$v = \cos(u), \quad dv = -\sin(u) \, du$

Thus, the integral becomes: $- \int \frac{dv}{v^3}$

Step 3: Solve the New Integral

The integral of $v^{-3}$ is straightforward: $\int v^{-3} \, dv = \frac{v^{-2}}{-2} = -\frac{1}{2v^2}$

Thus, the integral becomes: $-\left(-\frac{1}{2v^2}\right) = \frac{1}{2v^2}$

Step 4: Back Substitution

Since $v = \cos(u)$, we substitute back to get: $\frac{1}{2\cos^2(u)}$

Now, multiplying by the factor of 2 we had from earlier: $\frac{1}{\cos^2(u)}$

Step 5: Final Answer

Recalling that $u = \sqrt{x}$, we substitute back to get the final answer:

$\frac{1}{\cos^2(\sqrt{x})} + C$

Thus, the integral is:

$\boxed{\frac{1}{\cos^2(\sqrt{x})} + C}$

Would you like further clarification on any of these steps?

Here are 5 related questions to explore:

1. What is the integral of $\frac{1}{\cos^3(u)}$?
2. How does trigonometric substitution work in integrals involving sine and cosine?
3. How do you integrate functions involving square roots in the integrand?
4. What are some alternative methods for integrating trigonometric functions?
5. How can you recognize when to use substitution in integral problems?

Tip: When you see trigonometric functions with powers, think about identities or substitution to simplify the integration.