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The integral given is:

\[
\int \frac{1}{\sin^2 x \cos^2 x} \, dx
\]

To solve this, let's rewrite the integrand in a more manageable form.

1. We know that \(\sin^2 x \cos^2 x\) can be rewritten using a trigonometric identity:
   \[
   \sin^2 x \cos^2 x = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}
   \]

2. Therefore, we can rewrite the integral as:
   \[
   \int \frac{1}{\sin^2 x \cos^2 x} \, dx = \int \frac{4}{\sin^2 2x} \, dx
   \]

3. Since \(\frac{1}{\sin^2 2x} = \csc^2 2x\), we can rewrite this as:
   \[
   4 \int \csc^2 2x \, dx
   \]

4. Now, the integral of \(\csc^2 u\) with respect to \(u\) is \(-\cot u\). Here, let \(u = 2x\), so \(du = 2 \, dx\) or \(dx = \frac{du}{2}\).

5. Substitute into the integral:
   \[
   4 \int \csc^2 u \cdot \frac{du}{2} = 2 \int \csc^2 u \, du = 2(-\cot u) = -2 \cot(2x)
   \]

Thus, the solution to the integral is:

\[
\int \frac{1}{\sin^2 x \cos^2 x} \, dx = -2 \cot(2x) + C
\]

Would you like further explanation on any of these steps, or have questions about related concepts?

Here are five related questions for deeper understanding:

1. How would you integrate \(\int \csc^2 x \, dx\)?
2. What is the integral of \(\sec^2 x\) and how is it different from \(\csc^2 x\)?
3. How can trigonometric identities help simplify complex integrals?
4. How would you solve \(\int \frac{1}{\sin^4 x} \, dx\)?
5. What is the general method for integrating functions involving products of sine and cosine?

**Tip:** Using trigonometric identities can often simplify complex integrals, making them easier to evaluate.

Evaluate the integral \( \int \frac{1}{\sin^2 x \cos^2 x} \, dx \).

Learn how to solve the integral \( \int \frac{1}{\sin^2 x \cos^2 x} \, dx \) by simplifying it with trigonometric identities. This solution involves rewriting \( \sin^2(x) \cos^2(x) \) and integrating using substitution and the csc^2 identity. Suitable for undergraduate calculus students.

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