Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

C4 trigonometric integration

Announcements Posted on
Take our short survey, £100 of Amazon vouchers to be won! 23-09-2016
    • Thread Starter
    Offline

    0
    ReputationRep:
    is there a certain rule for integrating something like 2cos4x? When integrating cos x, I'm aware it goes to sin x + C. But isn't dividing by the coefficient of x or something involved when it comes to integrating something like 2cos4x?

    Thanks for any help.
    Offline

    0
    ReputationRep:
    There is a standard formula.
    Assuming K is a constant.
    Integral of coskx = 1/k x sinkx + C
    So your 2cos4x would be: 2x1/4xsin4x +C = 1/2sin4x
    Offline

    0
    ReputationRep:
    you mean

    $\int b\cos(ax)\;dx=b\cdot\frac{1}{a} sin(ax)+c

    Somit like that?
    • Thread Starter
    Offline

    0
    ReputationRep:
    thanks! gosh there's so many little rules, i tend to forget most of them!
    Offline

    1
    ReputationRep:
    Just use a simple substitution of u = 4x. hence du/dx = 4, and 1/4 du = dx

     \int 2cos(4x) dx = 2 \int cos(u) \frac{1}{4}du = 2.\frac{1}{4} \int cos(u) du = \frac{2}{4} sin(u) + C = \frac{1}{2}sin(4x) + C

    Just using substitution removes the need to memorise endless lists of general formulae.
    Offline

    2
    Sidhe's formula is correct, yep.

    Everytime you integrate a trig function like that, make sure that you differentiate it to check your answer.
    • Thread Starter
    Offline

    0
    ReputationRep:
    How would you integrate 2/sec^{2}x please?
    Offline

    3
    ReputationRep:
    \frac{2}{sec^2(x)} = 2cos^2(x)

    Use an identity to find 2cos^2(x) in terms of cos(2x).
    • Thread Starter
    Offline

    0
    ReputationRep:
    thanks!

    also, there was a cos(kx) formula given above, what's the sin(kx) one?
    Offline

    1
    ReputationRep:
    (Original post by cdeu12)
    thanks!

    also, there was a cos(kx) formula given above, what's the sin(kx) one?
    It's the same only it's cos instead of sin, and it's negative.
    Offline

    2
    ReputationRep:
    (Original post by notnek)
    \frac{2}{sec^2(x)} = 2cos^2(x)

    Use an identity to find 2cos^2(x) in terms of cos(2x).
    which identity would that be?
    Offline

    3
    ReputationRep:
    (Original post by M.A.H)
    which identity would that be?
    \displaystyle cos(2x)=2cos^2(x)-1
    Offline

    0
    ReputationRep:
    I'd just solve.

    $2\int\cos^2(x)\;dx

    But then I'm a weirdo.
    Offline

    2
    ReputationRep:
    (Original post by Sidhe)
    I'd just solve.

    $2\int\cos^2(x)\;dx
    And how are you planning on doing that? Using a trig identity is the only (quick) method of solving this integral.
    Offline

    0
    ReputationRep:
    (Original post by Hashshashin)
    And how are you planning on doing that? Using a trig identity is the only (quick) method of solving this integral.
    Like I said I'm a weirdo and I know what that equals off the top of my head. Believe it or not it's useful to know things like that, they may not be trig identitities per se but if you know them they save time.

    EDIT: actually I'm lying to be frank I already know what the derivative of the original is or at least 1/2 or d/dx of sec(x)^2 of it so in actuality, so I wouldn't of even bothered using anything, I'd of just written the answer out. After a while of doing problems you tend to get a hang of looking for simple methods that don't necessarily involve basic rules.

    Once you know how to derive the trig functions and have done a truck load there's no point in grinding out an answer, unless you want a bit of practice. That's what function tables and books are for, as long as you can derive them when you have to, you don't have to mess about. Of course this is absolutely no use or interest to the op, but then that's already answered.
    • Thread Starter
    Offline

    0
    ReputationRep:
    how would you integrate something like cos x sin^2 x?

    I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? Thanks.
    Offline

    3
    ReputationRep:
    (Original post by cdeu12)
    how would you integrate something like cos x sin^2 x?

    I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? Thanks.
    Recognition by noting that cosx is the derivative of sinx or use a substitution - u=sinx.
    • Thread Starter
    Offline

    0
    ReputationRep:
    thanks for the speedy reply notnek. i still don't really understand the recognition method, i'd be grateful it that could be explained a little more!
    Offline

    0
    ReputationRep:
    (Original post by cdeu12)
    how would you integrate something like cos x sin^2 x?

    I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? Thanks.
    By parts or substitution is most likely? What have you tried exactly? I'm not sure what you've done there?

    Try by parts is my advice.
    Offline

    3
    ReputationRep:
    (Original post by cdeu12)
    thanks for the speedy reply notnek. i still don't really understand the recognition method, i'd be grateful it that could be explained a little more!
    If you have something of the form:

    \displaystyle f'(x)(f(x))^{n}

    Then it's integral dx is \displaystyle \frac{(f(x))^{n+1}}{n+1} + c

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: April 20, 2008
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
How do you eat your pizza
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.