To find the integral of $\sin(\sqrt{x})$ with respect to $x$, we can proceed as follows:

$\int \sin(\sqrt{x}) \, dx$

Step 1: Substitution

Let's use the substitution method. Set: $u = \sqrt{x}$ Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{1}{2\sqrt{x}} \quad \text{or} \quad du = \frac{1}{2\sqrt{x}} \, dx$ Since $u = \sqrt{x}$, we can express $dx$ as: $dx = 2u \, du$

Step 2: Substitute and Simplify

Now, substitute $u$ and $dx$ into the integral: $\int \sin(\sqrt{x}) \, dx = \int \sin(u) \cdot 2u \, du$

Step 3: Integration by Parts

We'll use integration by parts, which is given by: $\int u \, dv = uv - \int v \, du$ Here, let: $u = u \quad \text{and} \quad dv = 2\sin(u) \, du$ Differentiate and integrate accordingly: $du = du \quad \text{and} \quad v = -2\cos(u)$

So, the integral becomes: $\int 2u \sin(u) \, du = -2u\cos(u) + 2\int \cos(u) \, du$

Step 4: Integrate $\cos(u)$

$\int \cos(u) \, du = \sin(u)$

Step 5: Combine the Results

Substitute back: $\int 2u \sin(u) \, du = -2u\cos(u) + 2\sin(u) + C$ Finally, replace $u$ with $\sqrt{x}$: $\int \sin(\sqrt{x}) \, dx = -2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C$

This is the final answer: $\boxed{\int \sin(\sqrt{x}) \, dx = -2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C}$

Would you like more details on any specific part, or do you have any questions?

Related Questions

1. How do you integrate $\sin(x^2)$?
2. What is the integral of $\cos(\sqrt{x})$?
3. Can you explain the integration by parts method in more detail?
4. How do you solve an integral using the substitution method?
5. What are some common integrals involving trigonometric functions?

Tip:

When performing integration, always consider if substitution or integration by parts can simplify the problem.