Show that $\dfrac{d}{dx}(\cot x) = -\csc^{2} x$ . Hence find the value of $\displaystyle \int_{\pi/4}^{\pi/2} -\cot^{2} x\, dx$ .

1. Express $\cot x$ in terms of $\sin x$ and $\cos x$ and apply the quotient rule.

$\cot x = \frac{\cos x}{\sin x}$

Let $u = \cos x$ and $v = \sin x$.
Then $\frac{du}{dx} = -\sin x$ and $\frac{dv}{dx} = \cos x$.

Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$:
$\frac{d}{dx}(\cot x) = \frac{\sin x (-\sin x) - \cos x (\cos x)}{(\sin x)^2}$

2. Simplify the numerator using the Pythagorean identity to obtain the final result.

$\frac{d}{dx}(\cot x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$
$= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$

Since $\sin^2 x + \cos^2 x = 1$:
$= \frac{-1}{\sin^2 x}$
$= -\csc^2 x$ (Shown)

1. Rewrite the integrand using a trigonometric identity to relate it to the result from part (a)

Using the identity $1 + \cot^2 x = \csc^2 x$, we have:
$-\cot^2 x = 1 - \csc^2 x$

2. Integrate the expression using the result from part (a)

$\displaystyle \int (1 - \csc^2 x)\, dx = x + \cot x$

3. Evaluate the definite integral at the upper and lower limits

$\displaystyle \int_{\pi/4}^{\pi/2} -\cot^2 x\, dx = \left[ x + \cot x \right]_{\pi/4}^{\pi/2}$
$= \left( \frac{\pi}{2} + \cot\left(\frac{\pi}{2}\right) \right) - \left( \frac{\pi}{4} + \cot\left(\frac{\pi}{4}\right) \right)$

4. Simplify to find the final exact value

$= \left( \frac{\pi}{2} + 0 \right) - \left( \frac{\pi}{4} + 1 \right)$
$= \frac{\pi}{4} - 1$