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Let's solve both integrals step by step.

### Problem 9: 
\[
\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx
\]

#### Solution:
Let’s perform a substitution to simplify this integral. Set:
\[
u = \sqrt{x} \quad \text{so} \quad x = u^2 \quad \text{and} \quad dx = 2u \, du.
\]

Thus, the integral becomes:
\[
\int \frac{\sin(u)}{u} \cdot 2u \, du = 2 \int \sin(u) \, du
\]
The integral of \(\sin(u)\) is:
\[
2 \int \sin(u) \, du = -2 \cos(u) + C
\]

Now substitute back \( u = \sqrt{x} \):
\[
\boxed{-2 \cos(\sqrt{x}) + C}
\]

---

### Problem 11:
\[
\int \cos^6(x) \sin^3(x) \, dx
\]

#### Solution:
We will use a reduction approach by separating powers of \(\sin(x)\) and \(\cos(x)\). Start by splitting \(\sin^3(x)\) as \(\sin^2(x) \cdot \sin(x)\) and using the identity \(\sin^2(x) = 1 - \cos^2(x)\).

So the integral becomes:
\[
\int \cos^6(x) \cdot (1 - \cos^2(x)) \sin(x) \, dx
\]
Now perform a substitution. Let:
\[
u = \cos(x) \quad \text{so} \quad du = -\sin(x) \, dx
\]
The integral becomes:
\[
-\int u^6 (1 - u^2) \, du = -\int (u^6 - u^8) \, du
\]

Now integrate each term:
\[
-\left( \frac{u^7}{7} - \frac{u^9}{9} \right) + C
\]
Substitute back \( u = \cos(x) \):
\[
\boxed{-\left( \frac{\cos^7(x)}{7} - \frac{\cos^9(x)}{9} \right) + C}
\]

---

Let me know if you need further explanation, and feel free to ask any questions! Here are five related problems you can explore:

1. Evaluate \( \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx \).
2. Solve \( \int \sin^2(x) \cos^4(x) \, dx \).
3. Find \( \int \cos^3(x) \sin^4(x) \, dx \).
4. Calculate \( \int \cos^2(x) \sin(x) \, dx \).
5. Work out \( \int \frac{1}{\sqrt{x^2 + 1}} \, dx \).

**Tip**: For trigonometric integrals involving powers of sine and cosine, substitution often simplifies the problem!

∫ (sin(√x) / √x) dx and ∫ cos^6(x) sin^3(x) dx

This page provides step-by-step solutions to two trigonometric integrals: ∫ (sin(√x) / √x) dx and ∫ cos^6(x) sin^3(x) dx. Learn how to use substitution and trigonometric identities to simplify and solve these integrals. Suitable for college-level calculus students.

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