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Sure, let's start with the first integral and proceed step by step.

### Problem 1:
Evaluate the integral \(\int \sin^4 x \cos^3 x \, dx\).

To solve this, we use the trigonometric identity and substitution method.

First, rewrite the integral:
\[ \int \sin^4 x \cos^3 x \, dx \]

Express \(\sin^4 x\) as \((\sin^2 x)^2\):
\[ \sin^2 x = 1 - \cos^2 x \]
Thus,
\[ \sin^4 x = (1 - \cos^2 x)^2 \]

Now, substitute \(u = \sin x\), which gives \(du = \cos x \, dx\):
\[ \int \sin^4 x \cos^3 x \, dx = \int (1 - \cos^2 x)^2 \cos^3 x \, dx \]

\[ = \int (1 - u^2)^2 u^3 \, du \]

Expand the integrand:
\[ = \int (1 - 2u^2 + u^4) u^3 \, du \]
\[ = \int (u^3 - 2u^5 + u^7) \, du \]

Now integrate term by term:
\[ = \int u^3 \, du - 2 \int u^5 \, du + \int u^7 \, du \]

\[ = \frac{u^4}{4} - 2 \cdot \frac{u^6}{6} + \frac{u^8}{8} + C \]

\[ = \frac{u^4}{4} - \frac{u^6}{3} + \frac{u^8}{8} + C \]

Substitute back \(u = \sin x\):
\[ = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{3} + \frac{\sin^8 x}{8} + C \]

So, the integral \(\int \sin^4 x \cos^3 x \, dx\) evaluates to:
\[ \frac{\sin^4 x}{4} - \frac{\sin^6 x}{3} + \frac{\sin^8 x}{8} + C \]

Would you like to proceed to the next problem or do you have any questions about this solution? 

Here are the next 5 problems for you to choose from:

1. \(\int \sin^5 x \, dx\)
2. \(\int \cos^5 x \, dx\)
3. \(\int x \cos^3 x^2 \sin x^2 \, dx\)
4. \(\int \sin^7 x \, dx\)
5. \(\int \cos^7 x \, dx\)

**Tip:** Always look for symmetry and substitution opportunities in trigonometric integrals to simplify your calculations.

Learn how to evaluate the integral of sin^4(x) cos^3(x) dx step-by-step using trigonometric identities and substitution method. This problem covers advanced integration techniques involving trigonometric functions. Suitable for undergraduate-level mathematics students.

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