This integral can be solved using trigonometric identities and substitution techniques. Let’s solve it step-by-step:

We aim to evaluate:

$\int \sin^4 x \cos^3 x \, dx$

Step 1: Break the powers and use substitution

To simplify, we split $\cos^3 x$ as: $\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x)\cos x$

So the integral becomes: $\int \sin^4 x \cos^3 x \, dx = \int \sin^4 x (1 - \sin^2 x) \cos x \, dx$

Step 2: Substitution

Let $u = \sin x$, which gives $du = \cos x \, dx$. The limits and powers of $\sin$ will update accordingly. Substituting, the integral becomes: $\int \sin^4 x (1 - \sin^2 x) \cos x \, dx = \int u^4 (1 - u^2) \, du$

Step 3: Expand and simplify

Expand $u^4(1 - u^2)$: $u^4(1 - u^2) = u^4 - u^6$

So the integral becomes: $\int u^4 - u^6 \, du$

Step 4: Integrate term by term

$\int u^4 \, du = \frac{u^5}{5}, \quad \int u^6 \, du = \frac{u^7}{7}$

Thus: $\int u^4 - u^6 \, du = \frac{u^5}{5} - \frac{u^7}{7} + C$

Step 5: Substitute back $u = \sin x$

Finally, substituting $u = \sin x$ back into the result gives: $\frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$

Final Answer:

$\boxed{\frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C}$

Would you like further clarification or details?

Here are some related questions for you to consider:

1. How do you decide when substitution is the best method for solving integrals?
2. Can you apply a similar method to solve $\int \sin^5 x \cos^3 x \, dx$?
3. What are the steps for integrating powers of trigonometric functions like $\sin^n x$ and $\cos^m x$?
4. How does the Pythagorean identity simplify trigonometric integrals?
5. How can integration by parts apply to trigonometric integrals?

Tip: When solving integrals involving even or odd powers of $\sin x$ or $\cos x$, consider substitution or using trigonometric identities strategically!