Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Somebody please help, I don't know how to do this!

    In = INT sec^n x dx

    (n-1)In = tanx sec^(n-2)x + (n-2)In-2

    Just a hint on how to start would be great, I'm at a loss.
    Offline

    0
    ReputationRep:
    Integrate by parts, choose u = sec^(n-2)x and dv/dx = sec^2(x).
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks
    Offline

    2
    ReputationRep:
    (Original post by queen-of-pain)
    Somebody please help, I don't know how to do this!

    In = INT sec^n x dx

    (n-1)In = tanx sec^(n-2)x + (n-2)In-2

    Just a hint on how to start would be great, I'm at a loss.

    I've only done up to P3 so this may be wrong, but I'm guessing:

    Int.sec^nx.dx=int.sec^2x.sec^(n-2)x.dx

    let u=sec^(n-2)x v'=sec^2x
    u'=(n-2)sec^(n-2)x.tanx v=tanx

    so int.sec^nx= sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
    = sec^(n-2)xtanx-int.(n-2)sec^(n-2)x(sec^2x-1).dx
    = sec^(n-2)xtanx-(n-2)int.sec^nx+(n-2)int.sec^(n-2)x.dx
    (n-1)int.sec^nx.dx=tanx.sec^(n-2)x+(n-2)int.sec^(n-2)x.dx

    which I think is the same thing.
    Offline

    2
    ReputationRep:
    (Original post by Ralfskini)
    I've only done up to P3 so this may be wrong, but I'm guessing:

    Int.sec^nx.dx=int.sec^2x.sec^(n-2)x.dx

    let u=sec^(n-2)x v'=sec^2x
    u'=(n-2)sec^(n-2)x.tanx v=tanx

    so int.sec^nx= sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
    = sec^(n-2)xtanx-int.(n-2)sec^(n-2)x(sec^2x-1).dx
    = sec^(n-2)xtanx-(n-2)int.sec^nx+(n-2)int.sec^(n-2)x.dx
    (n-1)int.sec^nx.dx=tanx.sec^(n-2)x+(n-2)int.sec^(n-2)x.dx

    which I think is the same thing.
    Excuse me, can anyone tell me whether the followig is acceptable:
    Int.sec^nx.dx
    =int.sec^2x.sec^(n-2)x.dx
    =int.sec^(n-2)x d(sec^2x) (<---- is it allowed?)
    sec^(n-2)xtanx-int.(n-2)sec^(n-2)xtan^2x.dx
    Offline

    0
    ReputationRep:
    Yes it is acceptable but in this case it is wrong. We have

    INT sec^nx dx
    = INT sec^(n-2)x (sec^2x dx)

    Now, you know that

    d(tanx)/dx = sec^2x

    Therefore,

    d(tanx) = sec^2x dx

    and then

    INT sec^(n-2)x (sec^2x dx) = INT sec^(n-2)x d(tanx)
    So, it is justified and I have seen it used once or twice in substitutions and when using parts. For example, say you have an integral with respect to x: INT f(x) dx and you apply some substitution, say u=x^2 which gives you an integral in some form INT g(u) du. You should remember that what you are actually doing now is integrating with repect to x^2 becuase u=x^2; that is, INT g(u) du = INT g(u) d(x^2). Also, for integration by parts we have

    INT u dv/dx dx = uv - INT v du/dx dx

    Which can actually be written

    INT u dv = uv - INT v du

    So, for example

    INT xe^x dx = INT x d(e^x)

    Comparing this to the above formula with u = x and v = e^x gives

    INT x d(e^x) = xe^x - INT e^x d(x) = (x-1)e^x + C

    So, the step is a valid one but I personally do not consider it to be a particularly practical or neat method.
 
 
 
Turn on thread page Beta
Updated: June 20, 2004
Poll
Do you think parents should charge rent?
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.