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

    0
    ReputationRep:
    I have to integrate the following with respect to x:

    cos2xsinx

    I found that when I used the substitution t = cosx i could then integrate the expression to obtain the correct answer but i was wondering if there was an easier way of solving this as usually the substition is given, but this time it was not.

    Thanx
    Offline

    3
    ReputationRep:
    (Original post by Nylex)
    cos 2x = 2cos^2 x - 1

    Int cos 2xsin x dx = [(2cos^2 x - 1)sin x] dx

    = (2sin xcos^2 x - sin x)dx

    For the first bit, use u = cos x, du = -sin x dx (take the 2 and the - outside the integral).
    GRRRRRRRRR YOU BEAT ME TO IT :P

    and I wanted 2 practise my P3 integration so I don't go rusty!
    Offline

    10
    ReputationRep:
    (Original post by Hoofbeat)
    GRRRRRRRRR YOU BEAT ME TO IT :P

    and I wanted 2 practise my P3 integration so I don't go rusty!
    Sorry. I've just deleted that post just now .
    Offline

    3
    ReputationRep:
    Why did you delete it Nylex?!
    Offline

    11
    ReputationRep:
    The way I'd do it is:

    I(x) = INT[cos2x.sinx] dx

    Using integration by parts,

    I(x) = ½sin2x.sinx - INT[½sin2x.cosx] dx

    Using parts again

    I(x) = ½sin2x.sinx - (-¼cos2x.cosx - INT[-¼cos2x.(-sinx)] dx)
    I(x) = ½sin2x.sinx - (-¼cos2x.cosx - ¼I(x))
    I(x) = ½sin2x.sinx + ¼cos2x.cosx + ¼I(x)
    ¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

    And then you can get I(x)...maybe there's some brilliant shortcut, but I can't see it right now.
    Offline

    10
    ReputationRep:
    (Original post by Hoofbeat)
    Why did you delete it Nylex?!
    Cos he said he used the substitution t = cos x and that's what I did. He wanted a different way of doing it.
    Offline

    3
    ReputationRep:
    Using Integration by parts you get:

    { means integration sign

    {cos2xsinx = -cosxcos2x - 2{sin2xcosx

    Now

    {sin2xcos = sin2xsinx - 2{cos2xsinx

    Therefore,

    {cos2xsinx = -cosxcos2x - 2[sin2xsinx - 2{cos2xsinx]
    {cos2xsinx = -cosxcos2x - 2sin2xsinx + 4{cos2xsinx
    {cos2xsinx - 4{cos2xsinx = -cosxcos2x - 2sin2xsinx
    -3{cos2xsinx = -cosxcos2x - 2sin2xsinx
    {cos2xsinx = (-1/3)[-cosxcos2x - 2sin2xsinx]


    {cos2xsinx = (1/3)[2sin2xsinx + cos2xcosx]

    I thinks thats right...
    Offline

    3
    ReputationRep:
    (Original post by Nylex)
    Cos he said he used the substitution t = cos x and that's what I did. He wanted a different way of doing it.
    oh rite. well I don't bother using the substitution (I know technically it is using the substitution) but I jsut recognise it's (=2sin xcos^2 x ) in the form f(x)^n.f'(x)
    Offline

    11
    ReputationRep:
    (Original post by integral_neo)
    Using Integration by parts you get:

    { means integration sign

    {cos2xsinx = -cosxcos2x + (1/2){sin2xcosx
    Wait, shouldn't that be -cosx.cos2x - INT[2cosx.sin2x] dx?
    Offline

    3
    ReputationRep:
    (Original post by Squishy)
    The way I'd do it is:

    I(x) = INT[cos2x.sinx] dx

    Using integration by parts,

    I(x) = ½sin2x.sinx - INT[½sin2x.cosx] dx

    Using parts again

    I(x) = ½sin2x.sinx - (-¼cos2x.cosx - INT[-¼cos2x.(-sinx)] dx)
    I(x) = ½sin2x.sinx - (-¼cos2x.cosx - ¼I(x))
    I(x) = ½sin2x.sinx + ¼cos2x.cosx + ¼I(x)
    ¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

    And then you can get I(x)...maybe there's some brilliant shortcut, but I can't see it right now.
    I dont think this is right...
    Offline

    10
    ReputationRep:
    (Original post by Hoofbeat)
    oh rite. well I don't bother using the substitution (I know technically it is using the substitution) but I jsut recognise it's (=2sin xcos^2 x ) in the form f(x)^n.f'(x)
    I don't think I was ever taught recognising f(x)^n.f'(x) and dealing with it like that.
    Offline

    11
    ReputationRep:
    (Original post by integral_neo)
    I dont think this is right...
    It is...I just checked.

    ¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

    Differentiate both sides:

    cos2x.sinx + ½sin2x.cosx - ½sin2x.cosx - ¼cos2x.sinx
    = ¾cos2x.sinx = ¾I'(x), as required

    since cos2x.sinx = I'(x)
    Offline

    15
    ReputationRep:
    (Original post by Nylex)
    I don't think I was ever taught recognising f(x)^n.f'(x) and dealing with it like that.
    Me either, theres no point memorising rules like that really, you can just do it in your head. You KNOW that the differentials of sinx and cosx are related so you can just fiddle about
    Offline

    10
    ReputationRep:
    (Original post by imasillynarb)
    Me either, theres no point memorising rules like that really, you can just do it in your head. You KNOW that the differentials of sinx and cosx are related so you can just fiddle about
    Works for me .
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Hoofbeat)
    GRRRRRRRRR YOU BEAT ME TO IT :P

    and I wanted 2 practise my P3 integration so I don't go rusty!
    You want some practice!! How would you go about integrating this:

    sin 2xcos x

    I used substitution but this method seems very long winded!
    Offline

    11
    ReputationRep:
    (Original post by the_anomaly)
    You want some practice!! How would you go about integrating this:

    sin 2xcos x

    I used substitution but this method seems very long winded!
    I'd use the wraparound method. Gets it done in a few lines.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Squishy)
    I'd use the wraparound method. Gets it done in a few lines.
    What's the wraparound method?? :confused:
    Offline

    11
    ReputationRep:
    (Original post by the_anomaly)
    What's the wraparound method?? :confused:
    Look at my posts above...if you apply it, you get the integral as

    -(sinx.sin2x + 2cosx.cos2x)/3
    Offline

    2
    ReputationRep:
    (Original post by the_anomaly)
    You want some practice!! How would you go about integrating this:

    ∫sin 2xcos x

    I used substitution but this method seems very long winded!
    ∫sin(2x)cosx
    ∫2sinxcosxcosx
    ∫2sinxcos²x
    -(2/3)cos³x
    =========
    Offline

    0
    ReputationRep:
    Integrate sec(x/2)tan(x/2).
    I got the answer 2sec(x/2), I think its wrong. anyone who can trace whats wrong?
 
 
 
Turn on thread page Beta
Updated: August 20, 2004
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.