Use the substitution formula to evaluate the integral. $$ \int_{\pi / 6}^{\pi / 2} \cot x \csc ^{3} x d x $$ $$ -\frac{7}{3} $$ $$ -\frac{15}{4} $$ $$ \frac{7}{3} $$ $$ \frac{15}{4} $$

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Let
$$
u = \csc x.
$$
Then
$$
\frac{du}{dx} = -\csc x \cot x \quad \Longrightarrow \quad -du = \csc x \cot x\,dx.
$$

Notice that the integrand is
$$
\cot x\, \csc^3 x\,dx = \csc^2 x \cdot (\cot x\,\csc x\,dx).
$$
Replacing $$\cot x\,\csc x\,dx$$ by $$-du$$ and noting that $$\csc^2 x = u^2$$, we have
$$
\cot x\, \csc^3 x\,dx = u^2\,(-du) = -u^2\,du.
$$

The original integral then becomes
$$
\int_{\pi/6}^{\pi/2} \cot x\, \csc^3 x\,dx = -\int_{u(\pi/6)}^{u(\pi/2)} u^2\,du.
$$

We now change the limits of integration. When
$$
x = \pi/6,\quad u = \csc(\pi/6) = 2,
$$
and when
$$
x = \pi/2,\quad u = \csc(\pi/2) = 1.
$$

Thus,
$$
-\int_{2}^{1} u^2\,du.
$$
Reversing the limits changes the sign:
$$
-\int_{2}^{1} u^2\,du = \int_{1}^{2} u^2\,du.
$$

We now evaluate the integral:
$$
\int_{1}^{2} u^2\,du = \left[\frac{u^3}{3}\right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.
$$

Thus, the value of the integral is
$$
\frac{7}{3}.
$$
The integral evaluates to $$ \frac{7}{3} $$.

Use the substitution formula to evaluate the integral.

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