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

P3 Help Please? Couple Questions watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    1. ∫ xsec ²x


    2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
    Offline

    1
    (Original post by Sang)
    1. ∫ xsec ²x


    2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
    1) ∫ xsec²x

    v=x
    dv/dx = 1
    du/dx=sec²x
    u=tanx

    => xtanx - ∫ tanx
    => xtanx - ln(secx)

    NB - The ln(secx) bit is from member when I would normally use the formula sheet so you might want to check what the integral of tanx actually is!!!
    Offline

    2
    ReputationRep:
    (Original post by Leekey)

    NB - The ln(secx) bit is from member when I would normally use the formula sheet so you might want to check what the integral of tanx actually is!!!
    yeah it is....because INT tanx = -ln(cosx) = ln(cosx)^-1 = ln(secx)
    Offline

    15
    ReputationRep:
    (Original post by Sang)
    1. ∫ xsec ²x


    2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
    question 2:

    y^2 dy = (x^4-1)/x^2 dx = x^2 - x^-2 dx

    so 1/3 y^3 = 1/3 x^3 + x^ -1 + C

    8/3 = 1/3 + 1 + C

    C = 4/3

    so y^3 = x^3 + 3x^-1 + 4
    Offline

    1
    (Original post by Sang)
    2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
    ∫ y² dy = ∫ (x^4-1)/(x²) dx

    You can take it from there...
    • Thread Starter
    Offline

    0
    ReputationRep:
    thanks people

    one question i dont understand is,

    ∫ xsec^x

    u=x
    du/dx=1
    dv/dx=sec^2x
    so v= lnsecxtanx

    so i did, xlnsecxtanx - ∫ lnsecxtanx

    then what? i was on the right track then wasn't i?
    Offline

    0
    ReputationRep:
    (Original post by Sang)
    dv/dx=sec^2x
    so v= lnsecxtanx
    v = tanx
    Offline

    15
    ReputationRep:
    (Original post by Sang)
    thanks people

    one question i dont understand is,

    ∫ xsec^x

    u=x
    du/dx=1
    dv/dx=sec^2x
    so v= lnsecxtanx

    so i did, xlnsecxtanx - ∫ lnsecxtanx

    then what? i was on the right track then wasn't i?
    umm if dv/dx is sec^2 x.... then v is tan x.

    so your integral should turn out to be x tanx - integral of tan x.
    Offline

    1
    ReputationRep:
    ∫ xsec²(x) dx

    f(x)=x ; g'(x)=sec²x=(tanx)'
    f'(x)=1 ; g(x)=tanx

    So, ∫ xsec²(x) = x*tanx - ∫1*tanx dx

    =xtanx - ∫sinx/cosx dx

    =xtanx + ∫(-sinx)/(cosx) dx

    =xtanx + ∫(cosx)'/cosx dx

    =xtanx + ln¦cosx¦ + k
    • Thread Starter
    Offline

    0
    ReputationRep:
    ahh excellent,

    my final question is this (more integration)

    a) Show that d/dx [e^3x (cos2x+ksin2x)] = e^3x[(3k-2)sin2x + (2k+3)cos2x]

    where k = constant

    b) Letting k=2/3 find ∫ e^3x cos2x dx

    c) Find ∫ e^3x sin2x dx


    then im sorted, thanks!
    Offline

    1
    ReputationRep:
    a) Show that d/dx [e^3x (cos2x+ksin2x)] = e^3x[(3k-2)sin2x + (2k+3)cos2x]
    differentiating using the product rule gives:

    = -2e^(3x).sin(2x) + 3e^(3x).cos(2x) + 3ke^(3x).sin(2x) + 2ke^(3x).cos(2x)

    = e^(3x).[sin(2x).(3k-2) cos(2x).(3+2k)]


    b) Letting k=2/3 find ∫ e^3x cos2x dx
    well when k = 2/3, you can substitute into the answer to (a) and then reverse the process in (a) to get the answer to the integral. Therefore substituting 2/3 gives the integral of e^(3x).cos2x and the answer is:

    e^(3x).cos(2x) + (2/3).e^(3x).sin(2x)


    c) Find ∫ e^3x sin2x dx
    now substitute k = -2/3 to remove cos(2x) term from the answer to (a). Then using the full method to (a) reversed, the answer is:

    e^(3x).sin(2x) - (2/3)e^(3x).sin(2x)

    Hope this helps.
 
 
 
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?

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.