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We are given the integral:

\[
\int \cot x \cdot \csc^3 x \, dx
\]

### Step-by-step Solution:

We can rewrite the integrand as:

\[
\cot x \cdot \csc^3 x = \csc^3 x \cdot \cot x
\]

This suggests that we could use substitution. Recall the derivative:

\[
\frac{d}{dx} (\csc x) = -\csc x \cdot \cot x
\]

Now, let:

\[
u = \csc x
\]

Then:

\[
du = -\csc x \cdot \cot x \, dx
\]

Thus, the integral becomes:

\[
\int \csc^2 x (-du) = -\int u^2 \, du
\]

This is a straightforward power rule integral:

\[
-\int u^2 \, du = -\frac{u^3}{3} + C
\]

Substitute back \( u = \csc x \):

\[
-\frac{\csc^3 x}{3} + C
\]

Thus, the result of the integral is:

\[
-\frac{\csc^3 x}{3} + C
\]

### Correct Option:
The correct answer is (d):

\[
-\frac{\csc^3 x}{3} + C
\]

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Would you like more details on this solution or have any questions?

Here are 5 related questions you could explore:
1. How do you compute the derivative of \( \csc x \) and \( \cot x \)?
2. What other types of integrals involve trigonometric identities like this one?
3. How can we use trigonometric substitution for more complex integrals?
4. Can this integral be solved using integration by parts?
5. How would you handle integrals involving powers of \( \sin x \) and \( \cos x \)?

**Tip**: Memorizing the derivatives and integrals of basic trigonometric functions like \( \sin x \), \( \cos x \), \( \tan x \), \( \csc x \), and \( \sec x \) can be very helpful in solving similar problems efficiently.

This solution guides you through solving the integral ∫ cot x . cosec^3 x dx using substitution. Core trigonometric identities and the substitution method simplify this integral, making it suitable for advanced high school students in Grades 11-12.

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