We will evaluate the integral $\int (\csc 2x)^2 \cot 2x \, dx$ using two different substitutions.

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(i) Using $u = \csc 2x$

1. Substitute $u = \csc 2x$: $\frac{du}{dx} = -2 \csc 2x \cot 2x \implies du = -2 \csc 2x \cot 2x \, dx$ Therefore: $\csc 2x \cot 2x \, dx = -\frac{1}{2} \, du$

2. Rewrite the integral: Substituting $u = \csc 2x$, we get: $\int (\csc 2x)^2 \cot 2x \, dx = \int u^2 \left(-\frac{1}{2} \, du\right) = -\frac{1}{2} \int u^2 \, du$

3. Integrate: $\int u^2 \, du = \frac{u^3}{3}$ Thus: $-\frac{1}{2} \int u^2 \, du = -\frac{1}{2} \cdot \frac{u^3}{3} = -\frac{u^3}{6}$

4. Substitute back $u = \csc 2x$: $-\frac{u^3}{6} = -\frac{(\csc 2x)^3}{6}$

So the solution is: $\boxed{-\frac{(\csc 2x)^3}{6} + C}$

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(ii) Using $u = \cot 2x$

1. Substitute $u = \cot 2x$: $\frac{du}{dx} = -2 (\csc 2x)^2 \implies du = -2 (\csc 2x)^2 \, dx$ Therefore: $(\csc 2x)^2 \, dx = -\frac{1}{2} \, du$

2. Rewrite the integral: Substituting $u = \cot 2x$, we get: $\int (\csc 2x)^2 \cot 2x \, dx = \int u \left(-\frac{1}{2} \, du\right) = -\frac{1}{2} \int u \, du$

3. Integrate: $\int u \, du = \frac{u^2}{2}$ Thus: $-\frac{1}{2} \int u \, du = -\frac{1}{2} \cdot \frac{u^2}{2} = -\frac{u^2}{4}$

4. Substitute back $u = \cot 2x$: $-\frac{u^2}{4} = -\frac{(\cot 2x)^2}{4}$

So the solution is: $\boxed{-\frac{(\cot 2x)^2}{4} + C}$

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Final Answers:

1. Using $u = \csc 2x$: $-\frac{(\csc 2x)^3}{6} + C$

2. Using $u = \cot 2x$: $-\frac{(\cot 2x)^2}{4} + C$

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Would you like detailed verification or further explanation? Here are five related questions for practice:

1. Integrate $\int (\csc x)^2 \cot x \, dx$ using substitution.
2. Solve $\int \cot x \, dx$ using a substitution of your choice.
3. Evaluate $\int (\csc x)^3 \, dx$.
4. Show how to derive the derivative of $\csc x$ using the quotient rule.
5. Prove the trigonometric identity $(\csc^2 x - 1) = \cot^2 x$.

Tip: For integrals involving $\csc x$ or $\cot x$, substitutions are often helpful for simplifying powers of trigonometric functions.