Let's walk through each integration step by step, using the most common methods available, without using integration by parts. We'll box the final answers as you requested.

1. $\int \sin^4(x) \, dx$

We start by expressing $\sin^4(x)$ as a combination of lower powers of sine using the power reduction formula:

$\sin^4(x) = \left( \sin^2(x) \right)^2 = \frac{1 - \cos(2x)}{2} \cdot \frac{1 - \cos(4x)}{2}$

Now, we simplify the expression:

$\sin^4(x) = \frac{1}{4} (1 - \cos(2x)) (1 - \cos(4x))$

Next, expand the product:

$= \frac{1}{4} (1 - \cos(2x) - \cos(4x) + \cos(6x))$

Now integrate term by term:

$\int \sin^4(x) \, dx = \frac{1}{4} \left( \int 1 \, dx - \int \cos(2x) \, dx - \int \cos(4x) \, dx + \int \cos(6x) \, dx \right)$

$= \frac{1}{4} \left( x - \frac{\sin(2x)}{2} - \frac{\sin(4x)}{4} + \frac{\sin(6x)}{6} \right) + C$

Thus, the final answer is:

$\boxed{ \frac{1}{4} \cos(4x) + \frac{1}{12} \cos(4x) + C }$

2. $\int \cos^4(x) \, dx$

Using a similar approach for $\cos^4(x)$:

$\cos^4(x) = \left( \cos^2(x) \right)^2 = \frac{1 + \cos(2x)}{2} \cdot \frac{1 + \cos(4x)}{2}$

Now expand the product:

$\cos^4(x) = \frac{1}{4} (1 + \cos(2x)) (1 + \cos(4x))$

After expanding:

$= \frac{1}{4} (1 + \cos(2x) + \cos(4x) + \cos(6x))$

Now integrate term by term:

$\int \cos^4(x) \, dx = \frac{1}{4} \left( x + \frac{\sin(2x)}{2} + \frac{\sin(4x)}{4} + \frac{\sin(6x)}{6} \right) + C$

Thus, the final answer is:

$\boxed{ \frac{1}{2} x - \frac{1}{4} \sin(2x) + \frac{1}{4} \sin(2x) + C }$