## Question 7:

### Part 7(a) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \cosh^n 2x\, dx = \int \cosh 2x \cosh^{n-1} 2x\, dx$ $= \frac{1}{2}\sinh 2x \cosh^{n-1} 2x - \int \frac{1}{2}\sinh 2x \times (n-1)\cosh^{n-2} 2x \times 2\sinh 2x\, dx$ | M1 A1 | M1: Correct split and attempts parts to obtain expression of correct form (sign and coefficient errors only). A1: Any correct expression |
| $= \frac{1}{2}\sinh 2x \cosh^{n-1} 2x - (n-1)\int \sinh^2 2x \cosh^{n-2} 2x\, dx$ Applies $\sinh^2 2x = \pm\cosh^2 2x \pm 1$ | dM1 | Requires previous M mark |
| $\Rightarrow I_n = \frac{1}{2}\sinh 2x \cosh^{n-1} 2x - (n-1)(I_n - I_{n-2})$ Introduces $I_n$ and $I_{n-2}$ | ddM1 | Not implied by given answer. Requires previous M mark |
| $\left\{\Rightarrow nI_n = \frac{1}{2}\sinh 2x \cosh^{n-1} 2x + (n-1)I_{n-2}\right\}$ $I_n = \frac{\sinh 2x \cosh^{n-1} 2x}{2n} + \frac{n-1}{n}I_{n-2}$ * | A1* | Fully correct proof. Condone missing $dx$'s. Poor bracketing must be recovered before given answer but no other errors e.g., sin for sinh, or wrong/missing arguments |

**Total: 5 marks**

### Part 7(a) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \cosh^n 2x\, dx = \int \cosh^2 2x \cosh^{n-2} 2x\, dx = \int (\sinh^2 2x + 1)\cosh^{n-2} 2x\, dx$ | M1 A1 | M1: Correct split and applies $\sinh^2 2x = \pm\cosh^2 2x \pm 1$ to obtain correct form (sign and coefficient errors only). A1: Correct expression |
| $\int \sinh^2 2x \cosh^{n-2} 2x\, dx \left\{= \int \sinh 2x \cosh^{n-2} 2x \sinh 2x\, dx\right\}$ $= \frac{1}{2(n-1)}\sinh 2x \cosh^{n-1} 2x - \frac{1}{n-1}\int \cosh^n 2x\, dx$ | dM1 | Attempts parts to obtain correct form for $\int \sinh^2 2x \cosh^{n-2} 2x\, dx$. Requires previous M mark |
| $\Rightarrow I_n = I_{n-2} + \frac{1}{2(n-1)}\sinh 2x \cosh^{n-1} 2x - \frac{1}{n-1}I_n$ Introduces $I_n$ and $I_{n-2}$ | ddM1 | Not implied by given answer. Requires previous M mark |
| $\left\{\Rightarrow (n-1)I_n = \frac{1}{2}\sinh 2x \cosh^{n-1} 2x + (n-1)I_{n-2} - I_n\right\}$ $I_n = \frac{\sinh 2x \cosh^{n-1} 2x}{2n} + \frac{n-1}{n}I_{n-2}$ * | A1* | Fully correct proof. Condone missing $dx$'s. Poor bracketing must be recovered. No other errors e.g., sin for sinh, or wrong/missing arguments |

**Total: 5 marks**

### Part 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1 + \cosh 2x)^3 = 1 + 3\cosh 2x + 3\cosh^2 2x + \cosh^3 2x$ | B1 | Correct expansion. Could be implied e.g. by $x + 3I_1 + 3I_2 + I_3$. Condone if partially or completely in "$x$" provided terms are collected |
| $\int \cosh^2 2x\, dx$ or $I_2 = \frac{1}{4}\sinh 2x \cosh 2x + \frac{1}{2}I_0$ or $\int \cosh^3 2x\, dx$ or $I_3 = \frac{1}{6}\sinh 2x \cosh^2 2x + \frac{2}{3}I_1$ | M1 | Completes attempt to apply reduction formula for $I_2$ or $I_3$. May be slips but must get two terms. May be seen with $I_0/I_1$ attempted and/or embedded in expression for $\int(1+\cosh 2x)^3\, dx$ |
| $I_0 = x$, $I_1 = \frac{1}{2}\sinh 2x$; $\int(1+3\cosh 2x)\, dx \to x \pm q\sinh 2x$ and uses above to obtain expression for $\int(1+\cosh 2x)^3\, dx$ | dM1 | $I_0 = x$ **and** $I_1 = \pm k\sinh 2x$ (condone $I_1$ from formula). Requires previous M mark |
| $= \frac{5}{2}x + \frac{11}{6}\sinh 2x + \frac{3}{4}\sinh 2x \cosh 2x + \frac{1}{6}\sinh 2x \cosh^2 2x\ (+c)$ | A1 | Correct answer with collected like terms seen |

**Total: 4 marks**

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