I see you've uploaded an image with three integrals. Let me break them down and provide solutions step by step:

1. Integral 1: $\int \cos(2x) \, dx$ Using the integral formula for cosines, we know: $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + C$ So, applying this for $k = 2$: $\int \cos(2x) \, dx = \frac{\sin(2x)}{2} + C_1$

2. Integral 2: $\int \sin(3x) \tan(3x) \, dx$ First, recall that $\tan(3x) = \frac{\sin(3x)}{\cos(3x)}$. Therefore, the integral becomes: $\int \sin(3x) \cdot \frac{\sin(3x)}{\cos(3x)} \, dx = \int \frac{\sin^2(3x)}{\cos(3x)} \, dx$ Now, let $u = \cos(3x)$, then $du = -3\sin(3x) \, dx$, which simplifies the integral: $-\frac{1}{3} \int \frac{1 - u^2}{u} \, du$ We can then solve this by breaking it into two simpler integrals.

3. Integral 3: $\int \sqrt{\cos^4(4x) + 2\sin^4(4x) - 1} \, dx$ Simplifying the terms inside the square root can be quite complex, and it might require algebraic manipulation or substitution methods depending on how to handle the expression efficiently.

Would you like me to continue with a more detailed breakdown for each integral or focus on a specific one?