Question 6 - A-Level Maths

CAIE FP1 2005 November — Question 6 8 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2005
Session	November
Marks	8
Paper	Download PDF ↗
Topic	Integration by Parts
Type	Reduction formula or recurrence
Difficulty	Challenging +1.2 This is a standard reduction formula question requiring integration by parts with a given hint. The differentiation hint makes finding the recurrence relation straightforward, and applying it twice with the known value I₁ = π/4 is routine. While it requires careful algebraic manipulation and is a multi-step problem worth 8 marks, the technique is a core Further Maths skill with no novel insight needed.
Spec	4.08c Improper integrals: infinite limits or discontinuous integrands

Let $$I_n = \int_0^1 (1 + x^2)^{-n} dx,$$ where $n \geqslant 1$. By considering $\frac{d}{dx}(x(1 + x^2)^{-n})$, or otherwise, prove that $$2nI_{n+1} = (2n - 1)I_n + 2^{-n}.$$ [5] Deduce that $I_3 = \frac{3}{32}\pi + \frac{1}{4}$. [3]

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Let
$$I_n = \int_0^1 (1 + x^2)^{-n} dx,$$
where $n \geqslant 1$. By considering $\frac{d}{dx}(x(1 + x^2)^{-n})$, or otherwise, prove that
$$2nI_{n+1} = (2n - 1)I_n + 2^{-n}.$$ [5]

Deduce that $I_3 = \frac{3}{32}\pi + \frac{1}{4}$. [3]

\hfill \mbox{\textit{CAIE FP1 2005 Q6 [8]}}

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