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

Test Ourselves: P3 watch

Announcements
    Offline

    0
    ReputationRep:
    right here it is....it's long and confusing and can be done better...but it works:


    x=sinu
    therefore 1-x^2 = cos^2 u
    dx/du = cos u
    dx = ducos u

    therefore INT sqrt(1-x^2) = INT cos^2 u

    integrate using cos^2u = .5(cos 2u +1)

    and you get:

    1/4 * sin 2u + 1/2 * u + c

    1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

    but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
    but now we simplify cos(arcsin x)
    substitue for x = sin u

    cos(arcsin sin u) = cos(u)
    cos^2(u) = 1-sin^2u
    so cos(u) = sqrt(1-sin^2 u)
    but sin u = x
    so
    cos(u) = cos(arcsin x) = sqrt(1-x^2)

    so:
    1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
    Offline

    1
    ReputationRep:
    (Original post by Silly Sally)
    Sorry i don't understand :confused: - what you did for the bold bit? Can you please explain
    (4xsin2x+2cos2x)/8

    4x(sin2x)= 4x(2sinxcosx) = 8xsinxcosx

    2(cos2x)=2(Cos^2x-Sin^2x) = 2((1-sin^2x)-sin^2x) = 2(1-2sin^2x) therefore:

    (4xsin2x+2cos2x)/8 = (8xsinxcosx - 4sin^2x + 2)/8
    Offline

    0
    ReputationRep:
    www.xs4all.nl/~mcgee/int.jpg

    that link should work.
    MrM;
    Offline

    0
    ReputationRep:
    www.xs4all.nl/~mcgee/Int.jpg


    that even
    • Thread Starter
    Offline

    0
    ReputationRep:
    what happens after u have substituted 1-sin^2 u for cos^2 u ....where does the square root go?
    Offline

    1
    ReputationRep:
    (Original post by scottyoneill)
    what happens after u have substituted 1-sin^2 u for cos^2 u ....where does the square root go?
    Taken out as a factor.
    Offline

    0
    ReputationRep:
    (Original post by Willa)
    right here it is....it's long and confusing and can be done better...but it works:


    x=sinu
    therefore 1-x^2 = cos^2 u
    dx/du = cos u
    dx = ducos u

    therefore INT sqrt(1-x^2) = INT cos^2 u

    integrate using cos^2u = .5(cos 2u +1)

    and you get:

    1/4 * sin 2u + 1/2 * u + c

    1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

    but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
    but now we simplify cos(arcsin x)
    substitue for x = sin u

    cos(arcsin sin u) = cos(u)
    cos^2(u) = 1-sin^2u
    so cos(u) = sqrt(1-sin^2 u)
    but sin u = x
    so
    cos(u) = cos(arcsin x) = sqrt(1-x^2)

    so:
    1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
    excellent! cheers!
    Offline

    0
    ReputationRep:
    (Original post by Willa)
    right here it is....it's long and confusing and can be done better...but it works:


    x=sinu
    therefore 1-x^2 = cos^2 u
    dx/du = cos u
    dx = ducos u

    therefore INT sqrt(1-x^2) = INT cos^2 u

    integrate using cos^2u = .5(cos 2u +1)

    and you get:

    1/4 * sin 2u + 1/2 * u + c

    1/4 * sin(2arcsinx) + 1/2 * arcsin x + c

    but sin(2arcsinx) = 2sin(arcsinx)cos(arcsinx) (double ange formula)
    but now we simplify cos(arcsin x)
    substitue for x = sin u

    cos(arcsin sin u) = cos(u)
    cos^2(u) = 1-sin^2u
    so cos(u) = sqrt(1-sin^2 u)
    but sin u = x
    so
    cos(u) = cos(arcsin x) = sqrt(1-x^2)

    so:
    1/4 * sqrt(1-x^2) + 1/2 * arcsin x + c
    I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
    Offline

    1
    ReputationRep:
    #jeepers creepers, where did you get those eyes from, I said jeepers creepers...#
    Offline

    14
    ReputationRep:
    (Original post by Silly Sally)
    I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
    no matter
    Offline

    2
    ReputationRep:
    Some slightly more challenging ones:

    ok, differentiate the following: x^(x^x)
    and (sqrtx)^(sqrtx).e^(x^2)

    and integrate these: cosxln(sinx)/sinx.dx

    ..and a toughy ((x^1/4)+5)/(x-16).
    Offline

    0
    ReputationRep:
    (Original post by Silly Sally)
    I am sure integrating arcsinx/getting arcsin x from an integral isn't on the edexcel sylabus for p3
    well you didnt have to integrate arcsin for that question, it was simply a matter of algebra.. rearranging x=sin u in terms of u, i.e. u = sin^1 x
    At least now you know!
    Offline

    0
    ReputationRep:
    (Original post by kimoni)
    well you didnt have to integrate arcsin for that question, it was simply a matter of algebra.. rearranging x=sin u in terms of u, i.e. u = sin^1 x
    At least now you know!
    Thanks - at least it won't annoy me the rest of the day!!!
    Offline

    1
    ReputationRep:
    (Original post by Ralfskini)
    Some slightly more challenging ones:

    ok, differentiate the following: x^(x^x)
    and (sqrtx)^(sqrtx).e^(x^2)

    and integrate these: cosxln(sinx)/sinx.dx

    ..and a toughy ((x^1/4)+5)/(x-16).
    Is this P3?
    • Thread Starter
    Offline

    0
    ReputationRep:
    yeh looks like it, the first 2 chain rule etc arent they?
    Offline

    2
    ReputationRep:
    (Original post by Bhaal85)
    Is this P3?

    It's well within the capabilities of a P3 student (I haven'd done beyond P3)
    Offline

    0
    ReputationRep:
    (Original post by Ralfskini)
    Some slightly more challenging ones:

    ok, differentiate the following: x^(x^x)
    and (sqrtx)^(sqrtx).e^(x^2)

    and integrate these: cosxln(sinx)/sinx.dx

    ..and a toughy ((x^1/4)+5)/(x-16).

    is the first one

    (x^x^x)(x + 2x*ln x) ?
    Offline

    0
    ReputationRep:
    (Original post by Ralfskini)
    Some slightly more challenging ones:

    ok, differentiate the following: x^(x^x)
    and (sqrtx)^(sqrtx).e^(x^2)

    and integrate these: cosxln(sinx)/sinx.dx

    ..and a toughy ((x^1/4)+5)/(x-16).
    cosxln(sinx)/sinx = cotxln(sinx)

    We have d(ln(sinx))/dx = cotx so a substitution of u=ln(sinx) will give

    INT cosxln(sinx)/sinx dx = (1/2)[ln(sinx)]^2 + C
    Offline

    2
    ReputationRep:
    (Original post by kimoni)
    is the first one

    (x^x^x)(x + 2x*ln x) ?

    No, keep trying.
    Offline

    2
    ReputationRep:
    (Original post by mikesgt2)
    cosxln(sinx)/sinx = cotxln(sinx)

    We have d(ln(sinx))/dx = cotx so a substitution of u=ln(sinx) will give

    INT cosxln(sinx)/sinx dx = (1/2)[ln(sinx)]^2 + C

    ...is the right answer!

    your method is probably better but I did it as follows:

    let u=sinx
    du=cosx.dx

    int.cosxlm(sinx)/sinx=int.(lnu/u).du

    Integrating by parts gives 1/2(lnu)^2+C.
 
 
 
Poll
Do I go to The Streets tomorrow night?

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.