The integral shown is:

$\int \frac{\cos^2(x) \, dx}{4\sin^2(x) - 1}$

Let's solve this step by step.

Step 1: Simplify the Denominator

Notice that the denominator $4\sin^2(x) - 1$ can be factored using the identity $a^2 - b^2 = (a - b)(a + b)$, with $a = 2\sin(x)$ and $b = 1$:

$4\sin^2(x) - 1 = (2\sin(x) - 1)(2\sin(x) + 1)$

So the integral becomes:

$\int \frac{\cos^2(x) \, dx}{(2\sin(x) - 1)(2\sin(x) + 1)}$

Step 2: Use a Trigonometric Identity for $\cos^2(x)$

We can apply the trigonometric identity $\cos^2(x) = 1 - \sin^2(x)$ to express the numerator in terms of $\sin(x)$:

$\cos^2(x) = 1 - \sin^2(x)$

Thus, the integral becomes:

$\int \frac{1 - \sin^2(x)}{(2\sin(x) - 1)(2\sin(x) + 1)} \, dx$

Step 3: Substitution

To simplify further, use the substitution $u = \sin(x)$, so that $du = \cos(x) \, dx$. Therefore, the integral now becomes:

$\int \frac{1 - u^2}{(2u - 1)(2u + 1)} \, du$

Now, we can break this fraction down using partial fraction decomposition, and solve the integral.

Shall I proceed with the partial fractions, or would you like to ask anything about the process so far?

Here are five relative questions:

1. How would the integral change if the denominator was $4\cos^2(x) - 1$ instead?
2. What are other examples of trigonometric substitutions that simplify integrals like this one?
3. Can this integral be solved using a different substitution method, such as $t = \tan\left(\frac{x}{2}\right)$?
4. How would the integral structure change if $\cos^2(x)$ in the numerator was $\sin^2(x)$ instead?
5. What alternative identities could simplify the expression if we had different powers of $\cos(x)$ or $\sin(x)$?

Tip: Trigonometric identities like $\cos^2(x) + \sin^2(x) = 1$ and factorization techniques often simplify complex integrals significantly.