Standard +0.8 This is a moderately challenging integration by substitution question requiring trigonometric substitution, use of the identity cos²θ = (1+cos2θ)/2, integration of cos2θ, careful limit transformation (x=1→θ=π/6, x=√3→θ=π/3), and verification of a specific exact answer. While the substitution is given, executing all steps correctly and arriving at the exact value π/3 requires solid technique and careful algebra—above average difficulty but within reach of competent A-level students.
Must see both surd values, or explanation as to why \(\sin\frac{2}{3}\pi = \sin\frac{2}{6}\pi\)
# Question 9:
| Answer | Mark | Guidance |
|--------|------|----------|
| $dx = 2\cos\theta \, d\theta$ | M1 (1.1a) | Attempt to link $dx$ and $d\theta$; allow sign error only |
| $\int\sqrt{4-x^2}\,dx = \int\sqrt{4-4\sin^2\theta}\cdot 2\cos\theta\,d\theta$ | M1 (3.1a) | Must substitute for both function and $dx$; can follow M0 but do not allow just $dx = d\theta$ |
| $= \int\sqrt{4\cos^2\theta}\cdot 2\cos\theta\,d\theta = \int 4\cos^2\theta\,d\theta$ | A1 (1.1) | Obtain correct integrand in terms of $\cos\theta$ only; condone no $d\theta$ |
| $= \int(2\cos 2\theta + 2)\,d\theta$ | M1 (2.1) | Attempt use of double angle formula; using $\cos 2\theta = \pm 2\cos^2\theta \pm 1$; integrand must be of form $k\cos^2\theta$ |
| $= \sin 2\theta + 2\theta$ | A1FT (1.1) | FT on $a\cos 2\theta + b$ only |
| $\left[\sin 2\theta + 2\theta\right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} = \left(\sin\frac{2}{3}\pi + \frac{2}{3}\pi\right) - \left(\sin\frac{2}{6}\pi + \frac{2}{6}\pi\right)$ | M1 (2.1) | Must be correct limits (either $x$ or $\theta$); correct order and subtraction; M0 for decimal values |
| $= \left(\frac{1}{2}\sqrt{3}+\frac{2}{3}\pi\right) - \left(\frac{1}{2}\sqrt{3}+\frac{1}{3}\pi\right)$ | | |
| $= \frac{1}{3}\pi$ **A.G.** | A1 (2.1) | Must see both surd values, or explanation as to why $\sin\frac{2}{3}\pi = \sin\frac{2}{6}\pi$ |
---