Reduction formulae — AS & A Level Further Mathematics (9231) questions by topic
Deriving and applying reduction formulae anchors this topic: finding the exact value of I_(-1) for an integral of (1 + x^2)^(n/2), showing a reduction formula relating I_n and I_(n-2) for powers of sech(x), and showing a similar reduction formula for an integral of (1-x)^n sinh(x).
Show and find are the two command words across 7 questions from 2023 to 2025 papers, worth from 7 to 11 marks, where integration by parts must be applied with exactly the right choice of parts to reveal the recursive pattern. With no examiner-report extracts logged for this topic, submitting each reduction-formula derivation for instant marking is the way to confirm the recursive relationship is proven cleanly before it is used to evaluate a specific term.
May–June 2025, Paper 21
Let In = integral from 0 to 1 of (1 - x)^n sinh(x) dx, where n is a non-negative integer. Show that, for n = 2, In = -1 + n(n - 1) I(n-2).
Answer this question and get it marked →May–June 2025, Paper 22
Let In = integral from 0 to 1 of (1 - x)^n sinh(x) dx, where n is a non-negative integer. Show that, for n = 2, In = -1 + n(n - 1)I{n-2}.
Answer this question and get it marked →May–June 2024, Paper 21
It is given that, for n = 0, In = integral from 0 to ln(3) of sech^n(x) dx. Show that, for n = 2, (n - 1)In = (3/5)^(n-2) (4/5) + (n - 2)I(n-2). [You may use the result that…
Answer this question and get it marked →May–June 2024, Paper 22
It is given that, for n = 0, In = integral from 0 to ln(3) of sech^n(x) dx. Show that, for n = 2, (n - 1)In = (3/5)^(n-2)(4/5) + (n - 2)I(n-2). [You may use the result that…
Answer this question and get it marked →May–June 2023, Paper 21
The integral In is defined by In = integral from 0 to 1 of (1 + x^5)^n dx. By considering d/dx (x(1 + x^5)^n), or otherwise, show that (5n + 1) In = 2^n + 5n I(n-1).
Answer this question and get it marked →May–June 2023, Paper 22
The integral In is defined by In = integral from 0 to 1 of (1 + x^5)^n dx. By considering (d/dx)(x(1 + x^5)^n), or otherwise, show that (5n + 1)In = 2^n + 5nI{n-1}.
Answer this question and get it marked →May–June 2023, Paper 23
The integral In, where n is an integer, is defined by In = integral from 0 to 4/3 of (1 + x^2)^(n/2) dx. Find the exact value of I(-1) giving your answer in the form ln(a), where…
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