4 Let \(I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { n } } \mathrm {~d} x\). Prove that, for every positive integer \(n\),
$$2 n I _ { n + 1 } = 2 ^ { - n } + ( 2 n - 1 ) I _ { n }$$
Given that \(I _ { 1 } = \frac { 1 } { 4 } \pi\), find the exact value of \(I _ { 3 }\).
Show mark scheme
Show mark scheme source
Question 4:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(I_n = \int_0^1 \frac{1}{(1+x^2)^n}\,dx = \left[\frac{x}{(1+x^2)^n}\right]_0^1 + \int_0^1 n(1+x^2)^{-(n+1)}\cdot2x^2\,dx\) M1A1
Integrates by parts
\(= 2^{-n} + 2n\int_0^1(1+x^2)^{-(n+1)}(1+x^2-1)\,dx\) M1A1
Rearranges
\(\therefore 2nI_{n+1} = 2^{-n}+(2n-1)I_n\) (AG) A1
Obtains result
\(2I_2 = \frac{1}{2}+\frac{1}{4}\pi \Rightarrow I_2 = \frac{1}{4}+\frac{1}{8}\pi\) M1A1
Uses reduction formula to find \(I_2\)
\(4I_3 = \frac{1}{4}+\frac{3}{4}+\frac{3}{8}\pi \Rightarrow I_3 = \frac{1}{4}+\frac{3}{32}\pi\) A1
Uses reduction formula to find \(I_3\)
Total: [8]
Copy
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \int_0^1 \frac{1}{(1+x^2)^n}\,dx = \left[\frac{x}{(1+x^2)^n}\right]_0^1 + \int_0^1 n(1+x^2)^{-(n+1)}\cdot2x^2\,dx$ | M1A1 | Integrates by parts |
| $= 2^{-n} + 2n\int_0^1(1+x^2)^{-(n+1)}(1+x^2-1)\,dx$ | M1A1 | Rearranges |
| $\therefore 2nI_{n+1} = 2^{-n}+(2n-1)I_n$ (AG) | A1 | Obtains result |
| $2I_2 = \frac{1}{2}+\frac{1}{4}\pi \Rightarrow I_2 = \frac{1}{4}+\frac{1}{8}\pi$ | M1A1 | Uses reduction formula to find $I_2$ |
| $4I_3 = \frac{1}{4}+\frac{3}{4}+\frac{3}{8}\pi \Rightarrow I_3 = \frac{1}{4}+\frac{3}{32}\pi$ | A1 | Uses reduction formula to find $I_3$ |
**Total: [8]**
---
Show LaTeX source
Copy
4 Let $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { n } } \mathrm {~d} x$. Prove that, for every positive integer $n$,
$$2 n I _ { n + 1 } = 2 ^ { - n } + ( 2 n - 1 ) I _ { n }$$
Given that $I _ { 1 } = \frac { 1 } { 4 } \pi$, find the exact value of $I _ { 3 }$.
\hfill \mbox{\textit{CAIE FP1 2013 Q4 [8]}}