Turn on thread page Beta

Integration problems watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    Lo, I'm having problems with these integration questions:

    1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

    If x = sinh u, dx = cosh u du

    => cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

    cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

    => cosh u du/cosh usinh^2 u

    => du/sinh^2 u

    Have I done the substitution right? If so, how do I integrate cosech^2 u?

    2. Parts: integral of (x^a)ln x dx (a is any real number)

    = uv - integral of v(du/dx)

    Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

    = (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

    = (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

    Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

    3. Parts again: ln [x + (1 + x^2)^1/2]

    Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

    If anyone can help, thanx a lot.
    Offline

    2
    ReputationRep:
    (Original post by Nylex)
    Lo, I'm having problems with these integration questions:

    1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

    If x = sinh u, dx = cosh u du

    => cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

    cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

    => cosh u du/cosh usinh^2 u

    => du/sinh^2 u

    Have I done the substitution right? If so, how do I integrate cosech^2 u?

    2. Parts: integral of (x^a)ln x dx (a is any real number)

    = uv - integral of v(du/dx)

    Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

    = (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

    = (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

    Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

    3. Parts again: ln [x + (1 + x^2)^1/2]

    Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

    If anyone can help, thanx a lot.
    yeeessss....no sorry i think only 'theone', elpaw or James F (in my year too!!) can help you here :-)
    Offline

    13
    ReputationRep:
    (Original post by Nylex)
    = (1/a + 1)x^(a+1).ln x - (1/a + 1)integral of x^(a + 1).1/x dx
    i don't know if that helps
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by elpaw)
    i don't know if that helps
    Oops. I've got that in my written solution, just missed it out when typing, lol.

    My answer for that one turns out to be:

    (1/a + 1)[x^(a+1)]ln x - [1/(a + 1)^2]x^(a + 1) + C

    It's definetly wrong .
    Offline

    13
    ReputationRep:
    (Original post by Nylex)
    Lo, I'm having problems with these integration questions:

    1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

    If x = sinh u, dx = cosh u du

    => cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

    cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh^2 u

    => cosh u du/cosh^2 u sinh^2 u
    => du/(cosh u sinh²u)
    Offline

    0
    ReputationRep:
    (Original post by Nylex)
    Lo, I'm having problems with these integration questions:

    1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

    If x = sinh u, dx = cosh u du

    => cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

    cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

    => cosh u du/cosh usinh^2 u

    => du/sinh^2 u

    Have I done the substitution right? If so, how do I integrate cosech^2 u?

    2. Parts: integral of (x^a)ln x dx (a is any real number)

    = uv - integral of v(du/dx)

    Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

    = (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

    = (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

    Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

    3. Parts again: ln [x + (1 + x^2)^1/2]

    Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

    If anyone can help, thanx a lot.
    tell me,wot do u get if u differentiate coth(x)...............
    Offline

    13
    ReputationRep:
    (Original post by IntegralAnomaly)
    tell me,wot do u get if u differentiate coth(x)...............
    coth(x)cosh(x)
    Offline

    0
    ReputationRep:
    (Original post by elpaw)
    coth(x)cosh(x)
    errr no u get -cosech^2(x), hmmm well if it isnt the integrand!
    Offline

    0
    ReputationRep:
    so the integral of cosech^2(u) is -coth(u)
    Offline

    13
    ReputationRep:
    (Original post by IntegralAnomaly)
    errr no u get -cosech^2(x), hmmm well if it isnt the integrand!
    sorry, was thinking of cosech (god all the h's are so confusing...)
    Offline

    0
    ReputationRep:
    (Original post by elpaw)
    sorry, was thinking of cosech (god all the h's are so confusing...)
    sooo true.....,whats worse is how u pronounce em.
    Offline

    13
    ReputationRep:
    (Original post by IntegralAnomaly)
    sooo true.....,whats worse is how u pronounce em.
    i al;ways pronounc them "sine haitch" "cos haitch" etc, and my teacher always had a go at me, saying they should be prononced "shine" and "cosh"
    Offline

    0
    ReputationRep:
    (Original post by elpaw)
    i al;ways pronounc them "sine haitch" "cos haitch" etc, and my teacher always had a go at me, saying they should be prononced "shine" and "cosh"
    u use to say sine-h-x? well it beats shine(x)
    Offline

    13
    ReputationRep:
    (Original post by IntegralAnomaly)
    u use to say sine-h-x? well it beats shine(x)
    yeah i did. whats wrong with that?
    Offline

    0
    ReputationRep:
    (Original post by elpaw)
    yeah i did. whats wrong with that?
    well for starters its not pronounced like that!
    Offline

    13
    ReputationRep:
    (Original post by IntegralAnomaly)
    well for starters its not pronounced like that!
    who says it isn't? who decided it shouold be 'shine'?
    Offline

    0
    ReputationRep:
    (Original post by elpaw)
    who says it isn't? who decided it shouold be 'shine'?
    the mathematician who invented it,anyway i dont blame u,ur a physicist.
    in physics u dont care wot its name is just how it works!
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by IntegralAnomaly)
    so the integral of cosech^2(u) is -coth(u)
    Thanx .

    Note: Apparently ln [x + (1 + x^2)^1/2] = sinh^-1 x, so I'm assuming I can just do the integral of that by parts (ie. 1sinh^-1 x)

    I've been told to pronounce the hyperbolic functions as shine and cosh as well, though I've heard people pronounce sinh x as sinesh x.
    Offline

    0
    ReputationRep:
    (Original post by Nylex)
    Lo, I'm having problems with these integration questions:

    2. Parts: integral of (x^a)ln x dx (a is any real number)

    3. Parts again: ln [x + (1 + x^2)^1/2]

    If anyone can help, thanx a lot.
    2) integral of (x^a) ln(x) = (1/(a+1))ln(x)(x^(a+1)) - integral of x^a/(a+1). This should be okay. This gives the initial integral to be (lnx . x^(a+1))/(a+1) - x^(a+1)/(a+1)^2. I think this is what you got?

    3) As you've said, just use parts on arsinh (x) .
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by theone)
    2) integral of (x^a) ln(x) = (1/(a+1))ln(x)(x^(a+1)) - integral of x^a/(a+1). This should be okay. This gives the initial integral to be (lnx . x^(a+1))/(a+1) - x^(a+1)/(a+1)^2. I think this is what you got?

    3) As you've said, just use parts on arsinh (x) .
    That is what I got. I couldn't differentiate last night and so thought it was wrong.

    Thanx .
 
 
 
Poll
Do you think parents should charge rent?

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.