## Question 8:

### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = \sec^{n-2}x$, $dv = \sec^2 x\,dx$; $du = (n-2)\sec^{n-3}x\cdot\sec x\tan x\,dx$, $v = \tan x$ | M1 | Attempt at integration by parts with correct $u$ and $v'$ |
| $I_n = \left[\sec^{n-2}x\tan x\right]_0^{\pi/4} - \int_0^{\pi/4}(n-2)\sec^{n-3}x\cdot\sec x\tan x\cdot\tan x\,dx$ | A1 | "$uv$" term must be seen |
| $= (\sqrt{2})^{n-2} - (n-2)\int_0^{\pi/4}\sec^{n-2}x\cdot\tan^2 x\,dx$ | A1 | Evaluating the first term |
| $= (\sqrt{2})^{n-2} - (n-2)\int_0^{\pi/4}\sec^{n-2}x(\sec^2 x - 1)\,dx$ | M1 | Dep on 1st M. Splitting integral using $\tan^2 x = \sec^2 x - 1$ |
| $= (\sqrt{2})^{n-2} - (n-2)(I_n - I_{n-2}) \Rightarrow (n-1)I_n = (\sqrt{2})^{n-2} + (n-2)I_{n-2}$ | A1 | [5] |

### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I_8 = \frac{1}{7}(2^3 + 6I_6) = \frac{1}{7}\left(2^3 + 6\cdot\frac{1}{5}(2^2+4I_4)\right)$ | B1 | Correct statement of formula seen anywhere |
| $= \frac{1}{7}\left(2^3 + \frac{6}{5}\left(2^2 + 4\cdot\frac{1}{3}(2+2I_2)\right)\right) = \frac{1}{7}\left(8 + \frac{6}{5}\cdot\frac{28}{3}\right)$ | B1 | Substitution of $I_2$ seen |
| $= \frac{1}{7}\cdot\frac{96}{5} = \frac{96}{35}$ | B1 | Conclusion. [3] |
| Alternative: $I_2=1$, $I_4=\frac{4}{3}$, $I_6=\frac{28}{15}$, $I_8=\frac{96}{35}$ | | B1 for $I_2$; B1 for $I_4$ and $I_6$; B1 |

### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I_2 = 1$ (which is rational); therefore $I_{2n}$ is rational for $n=1$ | B1 | Allow $n=2$ |
| Let $I_{2k}$ be rational for some value of $k$. Then $I_{2(k+1)} = \frac{1}{(2k+1)}\left(\sqrt{2}^{2k} + 2kI_{2k}\right) = \frac{1}{(2k+1)}\left(2^k + 2kI_{2k}\right)$ | M1 | For statement plus reduction formula and begin to look at $\sqrt{2}$ term; assume true for $n=k$ |
| Statement that this is rational; $\sqrt{2}^{2k} = 2^k$ is rational | A1 | Reduction formula must be correct |
| So if $I_{2k}$ is rational then $I_{2(k+1)}$ is rational. But $I_2$ is rational so $I_{2k}$ is rational for all $n=k$ | A1 | [4] |

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