Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Q14 in heinemann review exercise P5

    inn(between pi/4 and 0)=2/(3sin2x+4cos2x) dx
    the anwer =0.2ln6 but once more i can't get there!


    q11
    I think that there is a mistake/or i'm a twit!
    I(n)=int[between1,0]of x^(n-1)xroot(1-x^2) dx


    show that (n+2) I(n)=(n-1) I(n-2)

    I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!
    Offline

    0
    ReputationRep:
    For Q.14 you can get the answer with a t=tan(x) substitution. If you are not familiar with how to apply these feel free to write back and I will explain, also page 43 example 15 should help.

    Does anybody know of a better method for Q.14?
    Offline

    0
    ReputationRep:
    (Original post by mikesgt2)
    For Q.14 you can get the answer with a t=tan(x) substitution. If you are not familiar with how to apply these feel free to write back and I will explain, also page 43 example 15 should help.

    Does anybody know of a better method for Q.14?
    u could use 3sin2x+4cos2x=Rsin(2x+a).But i agree with using the t=tanx substitution.
    Offline

    1
    ReputationRep:
    (Original post by IntegralAnomaly)
    u could use 3sin2x+4cos2x=Rsin(2x+a).But i agree with using the t=tanx substitution.
    Thats what i first thought, but then what can you do??
    Offline

    0
    ReputationRep:
    (Original post by ogs)
    I think that the answer should be (n+1) at the beginning but anyones welcome to have a go!
    I get (n+2) on this question:

    u = x^(n-1), du/dx = (n-1)x^(n-2)
    dv/dx = x(1-x^2)^1/2, v = -(1/3)(1-x^2)^3/2

    Applying parts,

    I(n) = (n-1)/3 INT x^(n-2) (1-x^2)^3/2 dx
    3I(n) = (n-1) INT x^(n-2) (1-x^2)(1-x^2)^1/2 dx
    3I(n) = (n-1) INT [ x^(n-2) - x^n ](1-x^2)^1/2 dx
    3I(n) = (n-1)[ I(n-2) - I(n) ]

    So,

    3I(n) + (n-1)I(n) = (n-1)I(n-2)
    (n+2)I(n) = (n-1)I(n-2)
    Offline

    2
    ReputationRep:
    I just used the Rsin(2x+ø) method, but it's a bit tedious - which may explain why I got 0.5ln6 rather than 0.2ln6 !
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by mikesgt2)
    I get (n+2) on this question:

    u = x^(n-1), du/dx = (n-1)x^(n-2)
    dv/dx = x(1-x^2)^1/2, v = -(1/3)(1-x^2)^3/2

    Applying parts,

    I(n) = (n-1)/3 INT x^(n-2) (1-x^2)^3/2 dx
    3I(n) = (n-1) INT x^(n-2) (1-x^2)(1-x^2)^1/2 dx
    3I(n) = (n-1) INT [ x^(n-2) - x^n ](1-x^2)^1/2 dx
    3I(n) = (n-1)[ I(n-2) - I(n) ]

    So,

    3I(n) + (n-1)I(n) = (n-1)I(n-2)
    (n+2)I(n) = (n-1)I(n-2)

    thanks... I made a ridiculous mistake on the inital integration
    Offline

    0
    ReputationRep:
    here's the first:
    Attached Images
     
 
 
 
Turn on thread page Beta
Updated: June 18, 2004
Poll
Do you like carrot cake?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

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