| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Use trig identity before definite integration |
| Difficulty | Standard +0.3 This is a slightly-above-average A-level question requiring substitution to prove an integral identity, then using cos²x + sin²x = 1 to evaluate it, followed by a standard application of the double angle formula. The techniques are well-practiced but require careful execution across multiple parts. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
(i)(a) $du = -dx$ or equivalent [M1]
Substitute to obtain $\int_{\frac{1}{2}\pi}^{0} \cos^2(\frac{1}{2}\pi - u) \times (-du)$ [A1]
State $\cos(\frac{1}{2}\pi - u) = \sin u$ [A1]
State final result [AG] **3 marks**
(b) Common value $\frac{1}{4}\pi$ [B1] **1 mark**
(ii) State $\cos^2 x = \frac{1}{2}(1 + \cos 2x)$ [B1]
Attempt to integrate both terms [M1]
$\int_0^{\frac{1}{6}\pi} \cos^2 x\, dx = \frac{1}{2}\left[x + \frac{1}{2}\sin 2x\right]_0^{\frac{1}{6}\pi}$ [A1]
Substituting both limits [M1]
Answer $\frac{1}{12}\pi + \frac{1}{8}\sqrt{3}$ [A1] **5 marks**6
\begin{enumerate}[label=(\roman*)]
\item (a) Using the substitution $u = \frac { 1 } { 2 } \pi - x$, show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { 2 } x \mathrm {~d} x = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 2 } u \mathrm {~d} u$$
(b) Hence find the common value of these definite integrals.
\item Find the exact value of
$$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \cos ^ { 2 } x \mathrm {~d} x$$
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q6 [9]}}