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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? 

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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.

Learn how to solve the integral ∫ cos²(x) / (4sin²(x) - 1) dx using trigonometric identities and substitution. This detailed guide covers key concepts such as partial fractions and substitution methods for simplifying trigonometric integrals. Suitable for advanced calculus students.

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