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

    0
    ReputationRep:
    errr, no idea!
    Offline

    14
    \sin^3 x = \sin x \sin^2 x = \sin x (1- \cos^2 x).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Glutamic Acid)
    \sin^3 x = \sin x \sin^2 x = \sin x (1- \cos^2 x).
    can u carry on please...it's just a\bout confidence...i'd appreciate it.
    Offline

    14
    (Original post by FoOtYdUdE)
    can u carry on please...it's just a\bout confidence...i'd appreciate it.
    Expand it out, we have

    \displaystyle \int \sin x - \sin x \cos^2 x \, \text{d}x.

    You can integrate the first term without difficulty, as for the second: what is the derivative of cos x? Can you see how this will help you integrate it?
    Offline

    0
    ReputationRep:
    Hokey-koke.

    Your aim is to find:
    \int sin^3(x)\ \text{d}x

    If you re-write it as:
    \int \{ \sin(x) \times \sin^2(x) \}\ \text{d}x

    Then you can use the identity:
    \sin^2(x) \equiv 1 - \cos^2(x)

    So now you're trying to find:
    \int \{ \sin(x) - \sin(x)cos^2(x) \}\ \text{d}x

    Let us know how you get on with this.
    If you want a hint about how to do that second term, just ask!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Glutamic Acid)
    Expand it out, we have

    \displaystyle \int \sin x - \sin x \cos^2 x \, \text{d}x.

    You can integrate the first term without difficulty, as for the second: what is the derivative of cos x? Can you see how this will help you integrate it?
    -sinx? err and no :o:
    Offline

    0
    ReputationRep:
    (Original post by FoOtYdUdE)
    -sinx? err and no :o:
    Have you looked at integration by substitution?

    Spoiler:
    Show
    If so, then u = cos(x) is your substitution.

    So \int \{ - \sin(x) \times \cos^2(x) \}\ \text{d}x

    Can be written as \int -sin(x) \times u^2\ \text{d}x

    Also: \text{d}x=\frac{\text{d}u}{-\sin(x)}

    So \int \{ - \sin(x) \times \cos^2(x) \}\ \text{d}x = -\sin(x) u^2 \frac{\text{d}u}{-\sin(x)}

    Does this help you?


    Or does the following look familiar:
    f'(x) \times \{ f(x) \} ^n
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by placenta medicae talpae)
    Have you looked at integration by substitution?

    Or does the following look familiar:
    f'(x) \times \{ f(x) \} ^n
    int by substn. yeah dun that.
    • Thread Starter
    Offline

    0
    ReputationRep:
    oh im so silly..we can use parts for this quest.
    Offline

    0
    ReputationRep:
    (Original post by FoOtYdUdE)
    oh im so silly..we can use parts for this quest.
    Can do, yep

    If you'd like to do it by substitution (which is probably easier!), you could take a peek at the spoiler in my last post?
    Offline

    14
    ReputationRep:
    (Original post by FoOtYdUdE)
    oh im so silly..we can use parts for this quest.
    You'll be forever using parts. Just use the chain rule. The derivative of cos^3(x) is -3sin(x)cos^2(x) so the integral of sinxcos^2(x) is -1/3cos^3(x).
    Offline

    2
    ReputationRep:
    A nice solution would be to use reduction formula....I can't remember it for (sinx)^n it's something like

    nI_n = -cosx(sinx)^{n-1} + (n-1)I_{(n-2)}

    For this substitute in n=3
    Offline

    0
    ReputationRep:
    (Original post by JamesyB)
    A nice solution would be to use reduction formula....I can't remember it for (sinx)^n it's something like

    nI_n = -cosx(sinx)^{n-1} + (n-1)I_{(n-2)}

    For this substitute in n=3
    What is I?
    Offline

    1
    ReputationRep:
    (sin^3X)/(3cos^2x)
    this is the way I would do it, dunno if it's correct
    Offline

    1
    ReputationRep:
    (Original post by FoOtYdUdE)
    oh im so silly..we can use parts for this quest.
    The trick here is to spot derivatives. Try differentiating cos^3(x), what do you get? How does this relate to sin(x)cos^2(x)? How can you use your identities to manipulate this fact and use it?
    Offline

    2
    ReputationRep:
    (Original post by placenta medicae talpae)
    What is I?
    I is used to mean Integral i.e.

    I = \int^3_4{x^{2} dx}

    I_n is the integral you're working out where part of it is replaced by an n. In this case the squared part of the integral is changed to an n.

    I_n = \int^3_4{x^{n} dx}

    You can then work out a general form for integrating complicated functions (usually where integration by parts would be used multiple times) in terms of I_n.

    Sorry it's not very clear, only just learnt it at school so .... If anyone can explain it better then feel free
    Offline

    0
    ReputationRep:
    (Original post by JamesyB)
    I is used to mean Integral i.e.

    I = \int^3_4{x^{2} dx}

    I_n is the integral you're working out where part of it is replaced by an n. In this case the squared part of the integral is changed to an n.

    I_n = \int^3_4{x^{n} dx}

    You can then work out a general form for integrating complicated functions (usually where integration by parts would be used multiple times) in terms of I_n.

    Sorry it's not very clear, only just learnt it at school so .... If anyone can explain it better then feel free
    Oh right yeah, I get you.
    This kind of thing is useful when trying to find things like:
    \int e^x \sin(x)\ \text{d}x isn't it.
    Offline

    2
    ReputationRep:
    (Original post by placenta medicae talpae)
    Oh right yeah, I get you.
    This kind of thing is useful when trying to find things like:
    \int e^x \sin(x)\ \text{d}x isn't it.
    It is indeed
    Offline

    0
    ReputationRep:
    (Original post by JamesyB)
    It is indeed
    Cheers for that then; I had never come across that notation before!
    I'll be using that in future, will I
    Offline

    2
    ReputationRep:
    If you're not confident with integrating by recognition, then after the initial sin^3 x = (1 - cos^2 x)sin x, just substitute u = cos x.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 16, 2010
Poll
Do you like carrot cake?
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.