To solve the integral $\int \frac{1}{\sin^3(2x)} \, dx$, we'll use a trigonometric identity and substitution.

Step 1: Use Trigonometric Identity

We can express $\sin^3(2x)$ as $\sin(2x)$ times $\sin^2(2x)$. Using the identity $\sin^2(2x) = 1 - \cos^2(2x)$, we can rewrite the integral as:

$\int \frac{1}{\sin^3(2x)} \, dx = \int \frac{1}{\sin(2x) \cdot \sin^2(2x)} \, dx = \int \frac{1}{\sin(2x) \cdot (1 - \cos^2(2x))} \, dx$

Step 2: Use the Substitution Method

Let $u = \cos(2x)$, so that $du = -2\sin(2x) \, dx$, or $\frac{du}{-2} = \sin(2x) \, dx$.

Substituting into the integral:

$\int \frac{1}{\sin(2x) \cdot (1 - u^2)} \, dx = -\frac{1}{2} \int \frac{1}{1 - u^2} \, du$

Step 3: Solve the Integral Using Partial Fraction Decomposition

The integral $\int \frac{1}{1 - u^2} \, du$ can be solved by recognizing it as a standard form:

$\int \frac{1}{1 - u^2} \, du = \frac{1}{2} \ln \left|\frac{1+u}{1-u}\right| + C$

Step 4: Substitute Back

Substitute $u = \cos(2x)$ back into the expression:

$-\frac{1}{2} \cdot \frac{1}{2} \ln \left|\frac{1+\cos(2x)}{1-\cos(2x)}\right| + C = -\frac{1}{4} \ln \left|\frac{1+\cos(2x)}{1-\cos(2x)}\right| + C$

Step 5: Simplify the Expression

Using the identity $\frac{1+\cos(2x)}{1-\cos(2x)} = \cot^2(x)$, the integral simplifies to:

$-\frac{1}{4} \ln(\cot^2(x)) + C = -\frac{1}{2} \ln|\cot(x)| + C$

Thus, the final answer is:

$\int \frac{1}{\sin^3(2x)} \, dx = -\frac{1}{2} \ln|\cot(x)| + C$

Summary

The integral $\int \frac{1}{\sin^3(2x)} \, dx$ evaluates to $-\frac{1}{2} \ln|\cot(x)| + C$, where $C$ is the constant of integration.

Would you like a detailed explanation of any specific step? Here are some related problems you might want to try:

1. Solve $\int \frac{1}{\sin^2(x)} \, dx$.
2. Evaluate $\int \frac{1}{\sin^4(x)} \, dx$.
3. Solve $\int \frac{1}{\cos^3(x)} \, dx$.
4. Evaluate $\int \frac{\sin(x)}{\cos^2(x)} \, dx$.
5. Solve $\int \frac{1}{\tan(x)} \, dx$.

Tip: When dealing with integrals of trigonometric functions, try to simplify using identities and consider substitution methods.