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.