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Let's tackle each problem systematically.

### Problem 1: \( \int \sin^3 x \cos^2 x \, dx \)

To solve this, we can rewrite \( \sin^3 x \) as \( \sin^2 x \cdot \sin x \) and use the identity \( \sin^2 x = 1 - \cos^2 x \).

1. Rewrite: \( \int \sin^3 x \cos^2 x \, dx = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx \)
2. Substitute \( u = \cos x \), \( du = -\sin x \, dx \).
3. This transforms the integral into \( -\int (1 - u^2) u^2 \, du \).
4. Expand and integrate: \( -\int (u^2 - u^4) \, du \).
5. Solution after integration: \( -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \).

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### Problem 2: \( \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin^5 x \cos^3 x \, dx \)

This problem also requires trigonometric identities and substitution, similar to Problem 1.

1. Rewrite \( \sin^5 x \) as \( (\sin^2 x)^2 \sin x = (1 - \cos^2 x)^2 \sin x \).
2. Substitute \( u = \cos x \), \( du = -\sin x \, dx \).
3. Transform limits: when \( x = \frac{\pi}{2}, u = 0 \); when \( x = \frac{3\pi}{4}, u = -\frac{\sqrt{2}}{2} \).
4. Integral becomes \( -\int_{0}^{-\frac{\sqrt{2}}{2}} (1 - u^2)^2 u^3 \, du \).
5. Expand and integrate, then substitute back if needed.

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### Problem 3: \( \int \cos^5 x \sin^4 x \, dx \)

Rewrite \( \sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2 \) and proceed similarly to Problems 1 and 2.

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### Problem 4: \( \int \sin^6 x \cos^3 x \, dx \)

Use the identity for powers of sine and cosine, and apply a similar method of substitution as in the previous problems.

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### Problem 5: \( \int_0^{\frac{\pi}{2}} \cos^5 x \, dx \)

For odd powers of cosine, express \( \cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x \) and use substitution \( u = \sin x \).

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### Problem 6: \( \int \sin^3 (3x) \, dx \)

Rewrite \( \sin^3 (3x) \) as \( \sin^2 (3x) \sin (3x) \) and use identities and substitution.

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### Problem 7: \( \int (1 + \cos x)^2 \, dx \)

Expand \( (1 + \cos x)^2 = 1 + 2 \cos x + \cos^2 x \) and integrate each term separately.

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### Problem 8: \( \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx \)

Use the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 (2x) \) and simplify.

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### Problem 9: \( \int x \cos^2 x \, dx \)

Use integration by parts: let \( u = x \) and \( dv = \cos^2 x \, dx \).

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### Problem 10: \( \int_0^{\frac{\pi}{2}} \sin^4 x \cos^2 x \, dx \)

Use identities to rewrite \( \sin^4 x \) and proceed with a similar substitution.

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I will prepare detailed solutions for each step above. Let me know if you'd like further details on any specific problem, and I can walk you through each solution individually.

**Questions:**
1. Would you like a step-by-step breakdown of each integral?
2. Are you familiar with trigonometric identities used for simplifying powers?
3. Do you need help with integration by parts for problem 9?
4. Would you like more examples on substitution in integrals?
5. Are there any specific integrals you’d like to understand better?

**Tip:** When dealing with integrals of powers of sine and cosine, always look for identities that can simplify the expression, especially for powers above two.

Evaluate the following integrals:
1. ∫ sin^3(x) cos^2(x) dx
2. ∫ (from π/2 to 3π/4) sin^5(x) cos^3(x) dx
3. ∫ cos^5(x) sin^4(x) dx
4. ∫ sin^6(x) cos^3(x) dx
5. ∫ (from 0 to π/2) cos^5(x) dx
6. ∫ sin^3(3x) dx
7. ∫ (1 + cos(x))^2 dx
8. ∫ (from 0 to π/2) sin^2(x) cos^2(x) dx
9. ∫ x cos^2(x) dx
10. ∫ (from 0 to π/2) sin^4(x) cos^2(x) dx

Step-by-step solutions for evaluating integrals with powers of sine and cosine, including substitution and integration by parts methods. This worksheet covers advanced integration techniques ideal for Calculus II students.

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