Show that $\cot \alpha  = \tan \left( {\frac{\pi }{2} - \alpha } \right)$ for $0 < \alpha  < \frac{\pi }{2}$.

Hence find $\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha  < \frac{\pi }{2}}$.

EITHER

use of a diagram and trig ratios

eg,

$\tan \alpha  = \frac{O}{A} \Rightarrow \cot \alpha  = \frac{A}{O}$

from diagram, $\tan \left( {\frac{\pi }{2} - \alpha } \right) = \frac{A}{O}$     R1

OR

use of $\tan \left( {\frac{\pi }{2} - \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} - \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} - \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}$     R1

THEN

$\cot \alpha  = \tan \left( {\frac{\pi }{2} - \alpha } \right)$     AG

[1 mark]

$\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x}  = [\arctan x]_{\tan \alpha }^{\cot \alpha }$     (A1)

Note:     Limits (or absence of such) may be ignored at this stage.

$= \arctan (\cot \alpha ) - \arctan (\tan \alpha )$     (M1)

$= \frac{\pi }{2} - \alpha  - \alpha$     (A1)

$= \frac{\pi }{2} - 2\alpha$     A1

[4 marks]

This was generally well done.

This was generally well done. Some weaker candidates tried to solve part (b) through use of a substitution, though the standard result $\arctan x$ was well known. A small number used $\arctan x + c$ and went on to obtain an incorrect final answer.