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.