Edexcel AEA 2010 June — Question 5 12 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2010
SessionJune
Marks12
PaperDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeTrigonometric substitution to simplify integral
DifficultyChallenging +1.8 This AEA question requires executing a non-standard substitution (x = 1 + u^{-1}) with careful algebraic manipulation of nested radicals, then applying the result to definite integrals with trigonometric limits requiring half-angle formula knowledge. While technically demanding with multiple steps, the substitution is given and the path is clear, making it challenging but not requiring deep novel insight.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08h Integration by substitution

5. $$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$
  1. Use the substitution \(x = 1 + u ^ { - 1 }\) to show that $$I = - \left( \frac { x + 1 } { x - 1 } \right) ^ { \frac { 1 } { 2 } } + c$$
  2. Hence show that $$\int _ { \sec \alpha } ^ { \sec \beta } \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x = \cot \left( \frac { \alpha } { 2 } \right) - \cot \left( \frac { \beta } { 2 } \right) , \quad 0 < \alpha < \beta < \frac { \pi } { 2 }$$

5.

$$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $x = 1 + u ^ { - 1 }$ to show that

$$I = - \left( \frac { x + 1 } { x - 1 } \right) ^ { \frac { 1 } { 2 } } + c$$
\item Hence show that

$$\int _ { \sec \alpha } ^ { \sec \beta } \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x = \cot \left( \frac { \alpha } { 2 } \right) - \cot \left( \frac { \beta } { 2 } \right) , \quad 0 < \alpha < \beta < \frac { \pi } { 2 }$$
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2010 Q5 [12]}}