Turn on thread page Beta
    • Thread Starter

    i have been trying to do this questions but i cannot solve them. please help me! the questions are from the review exercise of the Heinemann Pure Mathematics 5 textbook.
    Question 50, part c and d. (I can do a and b ok), and
    Question 61, part a (i could probably do part b and c because the formula is given, but because i haven't been able to derive it i haven't gotten to solve b and c).
    Attached Images

    50. I(n) = [(n-1)/n] I(n-2)
    I(2n) = [(2n-1)/(2n)] I(2n-2)
    = [(2n-1)/(2n)] [(2n-3)/(2n-2)] I(2n-4)
    = [(2n-1)/(2n)] [(2n-3)/(2n-2)] [(2n-5)/(2n-4)] ... I(0)

    Now, I(0) = pi/2. So:
    I(2n) = [(2n-1)/(2n)] [(2n-3)/(2n-2)] [(2n-5)/(2n-4)] ... pi/2
    = [1/2^(n)] . [(2n-1)(2n-3)(2n-5)...]/[n(n-1)(n-2)...] . pi/2
    = [1/2^(n+1)] . [(2n-1)(2n-3)(2n-5)...]/(n!) . pi

    You can re-write (2n-1)(2n-3)... as (2n!)/[(2n-2)(2n-4)...] = (2n!)/[2^(n) (n!)]. So:
    I(2n) = [1/2^(n+1)] . (2n!)/[2^(n) (n!)²] . pi
    = [(2n!)pi]/[2^(2n+1) (n!)²], as required.

    (d) Do the same for I(2n-1), then multiply it by the expression for I(2n).

    61. I(n) = 01 1/(x²+a²)^n dx
    (a) Integrate by parts with:
    u = (x²+a²)^(-n), dv/dx=1
    du/dx = -2nx.(x²+a²)^(-n-1), v=x

    I(n) = [x/(x²+a²)^n]{0, 1} + 2n 01 x²/(x²+a²)^(n+1) dx
    = 1/(1+a²)^n + 2n 01 (x²+a²-a²)/(x²+a²)^(n+1) dx
    = 1/(1+a²)^n + 2n 01 (x²+a²)/(x²+a²)^(n+1) dx - 2na² 01 1/(x²+a²)^(n+1)
    = 1/(1+a²)^n + 2n 01 1/(x²+a²)^n dx - 2na².I(n+1)
    = 1/(1+a²)^n + 2n.I(n) - 2na².I(n+1)

    2na² I(n+1) = 1/(1+a²)^n + 2n I(n) - I(n)
    = 1/(1+a²)^n + (2n-1) I(n), as required.
    • Thread Starter

    i listend to some music and then came back to the problem and i solved it. here is the solution in two parts. thank you for the answer to 50.
    Attached Images
Turn on thread page Beta
Updated: June 12, 2005
The home of Results and Clearing


people online now


students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Will you be tempted to trade up and get out of your firm offer on results day?
Useful resources

Make your revision easier


Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here


How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.