5 Let
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$
where \(n \geqslant 0\). Use the fact that \(\tan ^ { 2 } x = \sec ^ { 2 } x - 1\) to show that, for \(n \geqslant 2\),
$$I _ { n } = \frac { 1 } { n - 1 } - I _ { n - 2 }$$
Show that \(I _ { 8 } = \frac { 1 } { 7 } - \frac { 1 } { 5 } + \frac { 1 } { 3 } - 1 + \frac { 1 } { 4 } \pi\).
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Question 5:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(I_n = \int_0^{\pi/4} \tan^n x\,dx = \int_0^{\pi/4} \tan^{n-2}x(\sec^2 x - 1)\,dx\) M1
Uses \(\tan^2 x = \sec^2 x - 1\)
\(= \int_0^{\pi/4}\tan^{n-2}x\sec^2 x\,dx - I_{n-2} = \left[\frac{\tan^{n-1}x}{n-1}\right]_0^{\pi/4} - I_{n-2}\) M1A1
Integrates
\(I_n = \frac{1}{n-1} - I_{n-2}\) (AG) A1
Obtains reduction formula
\(I_0 = \int_0^{\pi/4}1\,dx = \frac{\pi}{4}\) B1
Evaluates \(I_0\)
\(I_2 = \int_0^{\pi/4}(\sec^2 x - 1)\,dx = [\tan x - x]_0^{\pi/4} = 1 - \frac{\pi}{4}\) (B1)
\(I_2 = 1 - I_0 \quad I_4 = \frac{1}{3} - 1 + I_0\) M1A1
Uses reduction formula
\(I_6 = \frac{1}{5} - \frac{1}{3} + 1 - I_0 \quad I_8 = \frac{1}{7} - \frac{1}{5} + \frac{1}{3} - 1 + \frac{\pi}{4}\) (AG) A1
Alternative for first part:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(\frac{d}{dx}(\tan^{n-1}x) = (n-1)\tan^{n-2}x\sec^2 x = (n-1)\tan^{n-2}x(1+\tan^2 x) = (n-1)\tan^{n-2}x + (n-1)\tan^n x\) M1A1
Integrating between \(0\) and \(\frac{\pi}{4}\): \([\tan^{n-1}x]_0^{\pi/4} = (n-1)I_{n-2} + (n-1)\) M1
\(\Rightarrow 1 = (n-1)I_{n-2} + (n-1)I_n \Rightarrow I_n = \frac{1}{n-1} - I_{n-2}\) A1
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# Question 5:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $I_n = \int_0^{\pi/4} \tan^n x\,dx = \int_0^{\pi/4} \tan^{n-2}x(\sec^2 x - 1)\,dx$ | M1 | Uses $\tan^2 x = \sec^2 x - 1$ |
| $= \int_0^{\pi/4}\tan^{n-2}x\sec^2 x\,dx - I_{n-2} = \left[\frac{\tan^{n-1}x}{n-1}\right]_0^{\pi/4} - I_{n-2}$ | M1A1 | Integrates |
| $I_n = \frac{1}{n-1} - I_{n-2}$ (AG) | A1 | Obtains reduction formula |
| $I_0 = \int_0^{\pi/4}1\,dx = \frac{\pi}{4}$ | B1 | Evaluates $I_0$ |
| $I_2 = \int_0^{\pi/4}(\sec^2 x - 1)\,dx = [\tan x - x]_0^{\pi/4} = 1 - \frac{\pi}{4}$ | (B1) | |
| $I_2 = 1 - I_0 \quad I_4 = \frac{1}{3} - 1 + I_0$ | M1A1 | Uses reduction formula |
| $I_6 = \frac{1}{5} - \frac{1}{3} + 1 - I_0 \quad I_8 = \frac{1}{7} - \frac{1}{5} + \frac{1}{3} - 1 + \frac{\pi}{4}$ (AG) | A1 | |
**Alternative for first part:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\tan^{n-1}x) = (n-1)\tan^{n-2}x\sec^2 x = (n-1)\tan^{n-2}x(1+\tan^2 x) = (n-1)\tan^{n-2}x + (n-1)\tan^n x$ | M1A1 | |
| Integrating between $0$ and $\frac{\pi}{4}$: $[\tan^{n-1}x]_0^{\pi/4} = (n-1)I_{n-2} + (n-1)$ | M1 | |
| $\Rightarrow 1 = (n-1)I_{n-2} + (n-1)I_n \Rightarrow I_n = \frac{1}{n-1} - I_{n-2}$ | A1 | |
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5 Let
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$
where $n \geqslant 0$. Use the fact that $\tan ^ { 2 } x = \sec ^ { 2 } x - 1$ to show that, for $n \geqslant 2$,
$$I _ { n } = \frac { 1 } { n - 1 } - I _ { n - 2 }$$
Show that $I _ { 8 } = \frac { 1 } { 7 } - \frac { 1 } { 5 } + \frac { 1 } { 3 } - 1 + \frac { 1 } { 4 } \pi$.
\hfill \mbox{\textit{CAIE FP1 2011 Q5 [8]}}