Integral0^pi/A sqrt(1 + (4x)) dx

We use the double angle identity for cosine: $1 + \cos(2\theta) = 2\cos^2(\theta)$.
In our integral, we have $1 + \cos(4x)$. Let $2\theta = 4x$, which means $\theta = 2x$.
Substituting this into the identity gives:

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

The integral becomes:

$$\int_0^{\pi/A} \sqrt{2\cos^2(2x)} \, dx$$

$$\int_0^{\pi/A} \sqrt{2} \sqrt{\cos^2(2x)} \, dx = \int_0^{\pi/A} \sqrt{2} |\cos(2x)| \, dx$$

For this integral to be straightforward, we assume that $\cos(2x) \ge 0$ over the entire interval of integration $[0, \frac{\pi}{A}]$. This typically means that $2x$ lies in the interval $[0, \frac{\pi}{2}]$, which implies $x$ lies in $[0, \frac{\pi}{4}]$. Therefore, we assume $\frac{\pi}{A} \le \frac{\pi}{4}$, or $A \ge 4$.
Under this assumption, $|\cos(2x)| = \cos(2x)$.
So the integral simplifies to:

$$\int_0^{\pi/A} \sqrt{2} \cos(2x) \, dx$$

We can pull the constant $\sqrt{2}$ out of the integral:

$$\sqrt{2} \int_0^{\pi/A} \cos(2x) \, dx$$

The integral of $\cos(ax)$ is $\frac{1}{a}\sin(ax)$. So, the integral of $\cos(2x)$ is $\frac{1}{2}\sin(2x)$.

$$= \sqrt{2} \left[ \frac{1}{2}\sin(2x) \right]_0^{\pi/A}$$

$$= \frac{\sqrt{2}}{2} \left[ \sin\left(2 \cdot \frac{\pi}{A}\right) - \sin(2 \cdot 0) \right]$$ $$= \frac{\sqrt{2}}{2} \left[ \sin\left(\frac{2\pi}{A}\right) - \sin(0) \right]$$

Since $\sin(0) = 0$:

$$= \frac{\sqrt{2}}{2} \left[ \sin\left(\frac{2\pi}{A}\right) - 0 \right]$$ $$= \frac{\sqrt{2}}{2} \sin\left(\frac{2\pi}{A}\right)$$

The value of the integral is $\frac{\sqrt{2}}{2} \sin\left(\frac{2\pi}{A}\right)$.