(a)     Given that \(\alpha  > 1\), use the substitution \(u = \frac{1}{x}\) to show that

\[\int_1^\alpha  {\frac{1}{{1 + {x^2}}}{\text{d}}x = \int_{\frac{1}{\alpha }}^1 {\frac{1}{{1 + {u^2}}}{\text{d}}x} } .\]

(b)     Hence show that \(\arctan \alpha  + \arctan \frac{1}{\alpha } = \frac{\pi }{2}\).

(a)     \(u = \frac{1}{x} \Rightarrow {\text{d}}u = - \frac{1}{{{x^2}}}{\text{d}}x\)     M1

\( \Rightarrow {\text{d}}x = - \frac{{{\text{d}}u}}{{{u^2}}}\)     A1

\(\int_1^\alpha {\frac{1}{{1 + {x^2}}}{\text{d}}x = - \int_1^{\frac{1}{\alpha }} {\frac{1}{{1 + {{\left( {\frac{1}{u}} \right)}^2}}}\frac{{{\text{d}}u}}{{{u^2}}}} } \)     A1M1A1

Note: Award A1 for correct integrand and M1A1 for correct limits.

\( = \int_{\frac{1}{\alpha }}^1 {\frac{1}{{1 + {u^2}}}{\text{d}}u\,\,\,\,\,} \)(upon interchanging the two limits)     AG

(b)     \(\arctan x_1^\alpha  = \arctan u_{\frac{1}{\alpha }}^1\)     A1

\(\arctan \alpha - \frac{\pi }{4} = \frac{\pi }{4} - \arctan \frac{1}{\alpha }\)     A1

\(\arctan \alpha + \arctan \frac{1}{\alpha } = \frac{\pi }{2}\)     AG

[7 marks]

This question was successfully answered by few candidates. Both parts of the question prescribed the approach which was required – “use the substitution” and “hence”. Many candidates ignored these. The majority of the candidates failed to use substitution properly to change the integration variables and in many cases the limits were fudged. The logic of part (b) was missing in many cases.