Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    how do you do this?

    integrate (7x-x^2)/(2-x)(x^2+1) dx between 0 and 1?

    pk

    cheers in advance
    • Thread Starter
    Offline

    0
    ReputationRep:
    they want me to show its equal to (7/2)ln2 - pi/4, but i dont know how to integrate thi other than partial fractions - and its taking forever, andyone know a shortcut?
    Offline

    8
    ReputationRep:
    rewrite the numerator as

    -(1+x²) -7(2-x) + 15

    and you end up with frractions and also do partial fractions.

    its not tooo long!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Fermat)
    rewrite the numerator as

    -(1+x²) -7(2-x) + 15

    and you end up with frractions and also do partial fractions.

    its not tooo long!
    how'd you do that numerator thing - never been taught that; also, where do you go from there - the concurrent x^2 term in teh denominator is annoying me) - how do i deal with it?

    also, how do you integrate 15/(2-x)(x^2 +1) that you get at the end?:confused:

    must be a short cut:rolleyes:

    thanks, pk
    Offline

    13
    ReputationRep:
    (7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
    (7x-x²) = A(x²+1) + (Bx+C)(2-x)
    x = 2: 10 = 5A => A=2
    Compare constants: 0 = A + 2C => C=-1
    Compare x² coefficient: -1 = A - B => B=3

    (7x-x²) / (2-x)(x²+1) = 2/(2-x) + (3x-1)/(x²+1)

    ∫ 2/(2-x) + (3x-1)/(x²+1) dx
    = ∫ 2/(2-x)dx + ∫3x/(x²+1)dx - ∫1/(x²+1)dx
    = -2ln(2-x) + (3/2)ln(x²+1) - arctan(x) + c

    (edit... silly me )
    Offline

    8
    ReputationRep:
    (Original post by SsEe)
    (7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
    (7x-x²) = A(x²+1) + (Bx+C)(2-x)
    ...
    that's simpler!

    (Original post by SsEe)
    ...
    So ∫2/(2-x) + (3x-1)/(x²+1) dx
    = -2ln(2-x) -arctan(x) - (3/2)ln(x²+1) + c
    That's what I got
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by SsEe)
    (7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
    (7x-x²) = A(x²+1) + (Bx+C)(2-x)
    x = 2: 10 = 5A => A=2
    Compare constants: 0 = A + 2C => C=-1
    Compare x² coefficient: -1 = A - B => B=3

    (7x-x²) / (2-x)(x²+1) = 2/(2-x) + (3x-1)/(x²+1)

    ∫ 2/(2-x) + (3x-1)/(x²+1) dx
    = ∫ 2/(2-x)dx + ∫3x/(x²+1)dx - ∫1/(x²+1)dx
    = -2ln(2-x) + (3/2)ln(x²+1) - arctan(x) + c

    (edit... silly me )
    did that but got my constants wrong, lol - no wonder i was going wrong
    • Thread Starter
    Offline

    0
    ReputationRep:
    stuck on another few questions!!!

    integral of sec^n dx between 0 and pi/3.. The resuction formula i got was

    In = (2n-2) .(sqrt 3)/n-1 - (n-1)/(n-2) . In-2

    Had to work out I7 - i keep on getting 103sqrt3 / 24 - (5/16) ln(2+sqrt 3), but the book has 61/8 rt 4 frp sp,e reason.

    help

    also stuck on these:
    integrate 1/(x+sqrt(1-x2) dx, using x=sint
    finally integrate sinx.(1+cos x) / (1-cosx) dx between pi/3 and pi/2
    thanks

    pk
    Offline

    2
    ReputationRep:
    For the last one, maybe try a t = tan(x/2) substitution...

    I've kinda rushed it. But if I've done it right, then you get: (4) INT 1/[t(t2+1)] dt, which you could do partials on, and see how it goes.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by mockel)
    For the last one, maybe try a t = tan(x/2) substitution...

    I've kinda rushed it. But if I've done it right, then you get: (4) INT 1/[t(t2+1)] dt, which you could do partials on, and see how it goes.
    1/[t(t2+1)] - kept on getting ->1/[t(t2+1)2]

    which i couldmn't integrate - i'll check my working again

    another question i'm having probs with is to show

    integral sin n x / sin x

    In= 2sin(n-1)x / (n-1) +In-2

    dont know what step to start with - -tried compound angle formula and taking the sin up and parts, but not getting anywhere - hope they give us and easy integrate (xn sin x)dx one in the exam
    Offline

    2
    ReputationRep:
    Yeah, sorry, it definitely is what you said. I'll think about it some more.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by mockel)
    Yeah, sorry, it definitely is what you said. I'll think about it some more.
    what a weird question huh- out of all the papers i've dont i've never come across such awkward integrals:rolleyes: - these are from the exercises in the book - never touched um till yesterday

    pk
    Offline

    2
    ReputationRep:
    The sinnx one, I remember doing. Didn't they say something like 'by considering I2 - In'?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by mockel)
    The sinnx one, I remember doing. Didn't they say something like 'by considering I2 - In'?
    if you gopt P5 edexcell - its from pg 55 at the top - doesn't give any hints - just say to show its that reduction thing,

    pk
    Offline

    10
    ReputationRep:
    For the sin(nx)/sinx one, note that:
    sin(nx) - sin[(n-2)x] = 2cos[(n-1)x]sinx

    Therefore:
    I(n) = ∫ (2cos[(n-1)x]sinx + sin[(n-2)x])/sinx) dx
    = ∫ 2cos[(n-1)x] dx + ∫ sin[(n-2)x]/sinx dx
    = [2sin(n-1)x]/(n-1) + I(n-2)

    integrate 1/(x+sqrt(1-x2) dx, using x=sint
    This is horrible! Are you sure you typed it out right?
    • Thread Starter
    Offline

    0
    ReputationRep:
    it was integral 1/(x+sqrt(1-x2)) using x=sin t, lol

    dont know how its supposed to help; i got integral of cos t/(sint + cos t), and didn't know where to progress from there:confused:

    pk
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by dvs)
    For the sin(nx)/sinx one, note that:
    sin(nx) - sin[(n-2)x] = 2cos[(n-1)x]sinx
    could you explain that line???

    how the hell are we suppoed to know that:confused:, lol - would never have thought of that in an exam - did you know it from practice, or are you a wizz with trig identities:cool:?

    figured out you used sum-product, but HOW did you spot that - starting to get worried about tomorrow - may be harder than i first though:rolleyes:

    pk
    Offline

    2
    ReputationRep:
    Crazy. Anyway, I remember the hint from the paper now, and it was by considering In-2 - In:

    In-2 - In = ∫ [sin(n+2)x - sin(n)x / sinx] dx

    Using the identity, sinA - sinB = 2cos[(A+B)/2]sin[(A-B)/2]

    In-2 - In = ∫ 2[cos(n+1)x.sinx] / sinx dx

    In-2 - In = (2) ∫ cos(n+1)x dx

    In-2 - In = 2sin(n+1)x/(n+1)

    => In = In-2 - 2sin(n+1)x/(n+1)


    Argh, too late. Should've done it the other way as well. If I delete it, it'll seem like a waste of time. So, I won't. =D
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by mockel)
    Crazy. Anyway, I remember the hint from the paper now, and it was by considering In-2 - In:

    In-2 - In = ∫ [sin(n+2)x - sin(n)x / sinx] dx

    Using the identity, sinA - sinB = 2cos[(A+B)/2]sin[(A-B)/2]

    In-2 - In = ∫ 2[cos(n+1)x.sinx] / sinx dx

    In-2 - In = (2) ∫ cos(n+1)x dx

    In-2 - In = 2sin(n+1)x/(n+1)

    => In = In-2 - 2sin(n+1)x/(n+1)

    Well isn't that strange. Probably made a mistake somewhere along the line...

    Argh, too late. Should've done it the other way as well. If I delete it, it'll seem like a waste of time. So, I won't. =D
    its n-2 - did they give that in the paper? - in the textbook, there are no hints, and i've no idea how we are supposed to 'see that'
    Offline

    2
    ReputationRep:
    Definitely gave the hint in the paper.
 
 
 
Turn on thread page Beta
Updated: June 20, 2005
The home of Results and Clearing

3,016

people online now

1,567,000

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
Poll
Do you want your parents to be with you when you collect your A-level results?
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.