# Question 4:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \ln\left(\frac{x}{2}\right)dx = \int 1\cdot\ln\left(\frac{x}{2}\right)dx$, with $u = \ln\left(\frac{x}{2}\right) \Rightarrow \frac{du}{dx} = \frac{1}{x}$; $\frac{dv}{dx} = 1 \Rightarrow v = x$ | | Setup for integration by parts |
| $\int \ln\left(\frac{x}{2}\right)dx = x\ln\left(\frac{x}{2}\right) - \int x\cdot\frac{1}{x}\,dx$ | M1 | Use of IBP formula in correct direction |
| Correct expression: $x\ln\left(\frac{x}{2}\right) - \int x\cdot\frac{1}{x}\,dx$ | A1 | Correct expression |
| $= x\ln\left(\frac{x}{2}\right) - \int 1\,dx$ | dM1 | Attempt to multiply $x$ by $\frac{a}{x}$ or $\frac{1}{bx}$ or $\frac{1}{x}$ |
| $= x\ln\left(\frac{x}{2}\right) - x + c$ | A1 aef | Correct integration with $+c$ |

**[4 marks]**

### Aliter Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\ln\left(\frac{x}{2}\right)dx = \int(\ln x - \ln 2)dx = \int\ln x\,dx - \int\ln 2\,dx$ | | Split of logarithm |
| $\int\ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx$ | M1 | IBP formula in correct direction |
| $= x\ln x - x + c$ | A1 | Correct integration of $\ln x$ with or without $+c$ |
| $\int\ln 2\,dx = x\ln 2 + c$ | M1 | Correct integration of $\ln 2$ with or without $+c$ |
| $\int\ln\left(\frac{x}{2}\right)dx = x\ln x - x - x\ln 2 + c$ | A1 aef | Correct integration with $+c$ |

**[4 marks]**

### Aliter Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \frac{x}{2} \Rightarrow \frac{du}{dx} = \frac{1}{2}$, so $\int\ln\left(\frac{x}{2}\right)dx = 2\int\ln u\,du$ | | Substitution applied correctly |
| $\int\ln u\,du = u\ln u - \int u\cdot\frac{1}{u}\,du$ | M1 | IBP formula in correct direction |
| $= u\ln u - u + c$ | A1 | Correct integration of $\ln u$ |
| $2\int\ln u\,du$ step | M1 | *Decide to award 2nd M1 here* |
| $\int\ln\left(\frac{x}{2}\right)dx = 2(u\ln u - u) + c = x\ln\left(\frac{x}{2}\right) - x + c$ | A1 aef | Correct integration with $+c$ |

**[4 marks]**

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## Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $\cos 2x = \pm1 \pm 2\sin^2 x$ or $\sin^2 x = \frac{1}{2}(\pm1 \pm \cos 2x)$ | M1 | Consideration of double angle formula for $\cos 2x$ |
| $= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2}\,dx = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(1-\cos 2x)\,dx$ | | Correct use of identity |
| $= \frac{1}{2}\left[x - \frac{1}{2}\sin 2x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$ | dM1, A1 | Integrating to give $\pm ax \pm b\sin 2x$, $a,b\neq 0$; correct result equivalent to $\frac{1}{2}x - \frac{1}{4}\sin 2x$ |
| $= \frac{1}{2}\left[\left(\frac{\pi}{2} - \frac{\sin(\pi)}{2}\right) - \left(\frac{\pi}{4} - \frac{\sin\left(\frac{\pi}{2}\right)}{2}\right)\right]$ | ddM1 | Substitutes limits of $\frac{\pi}{2}$ and $\frac{\pi}{4}$ and subtracts correct way round |
| $= \frac{1}{2}\left[(\frac{\pi}{2}-0)-(\frac{\pi}{4}-\frac{1}{2})\right]$ | | |
| $= \frac{1}{2}\left(\frac{\pi}{4}+\frac{1}{2}\right) = \frac{\pi}{8}+\frac{1}{4}$ | A1 **aef, cso** | $\frac{1}{2}\left(\frac{\pi}{4}+\frac{1}{2}\right)$ or $\frac{\pi}{8}+\frac{1}{4}$ or $\frac{\pi}{8}+\frac{2}{8}$; candidate must collect $\pi$ term and constant term together; no fluked answers |

**[5 marks]**

### Aliter Way 2 (IBP):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = \int\sin^2 x\,dx$; $u=\sin x$, $\frac{dv}{dx}=\sin x \Rightarrow v=-\cos x$ | | Setup |
| $I = \{-\sin x\cos x + \int\cos^2 x\,dx\}$ | M1 | Attempt to use correct IBP formula |
| $I = \{-\sin x\cos x + \int(1-\sin^2 x)dx\}$ | | Use of $\cos^2 x = 1-\sin^2 x$ |
| $2I = \{-\sin x\cos x + \int 1\,dx\}$ | dM1 | LHS becoming $2I$ |
| $\int\sin^2 x\,dx = \{-\frac{1}{2}\sin x\cos x + \frac{x}{2}\}$ | A1 | Correct integration |
| Substituting limits $\frac{\pi}{2}$ and $\frac{\pi}{4}$, subtracting correctly | ddM1 | |
| $= \frac{\pi}{8}+\frac{1}{4}$ | A1 **aef, cso** | No fluked answers |

**[5 marks]**

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