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Quick Maths Problem

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  • 25-03-2009 8:33pm
    #1
    Scoobydooo
    Registered Users Posts: 341 ✭✭


    Ran into this question when going through the papers its on integration

    evaluate: square root of 4 - x2 letting x = 2SINQ


    Limits are 0 and root 3

    the 2 in x2 is meant to be the squared sign and Q stands for theta

    Thanks in advance !!


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    #2
    andyman
    Closed Accounts Posts: 3,641 ✭✭✭andyman


    It's one of the earlier years isn't it?

    Circular integration that sounds like. Can't quite remember how to do them but you have to let u = 2sinθ then differentiate, I think.

    Could be wrong. If you use Aidan Roantrees book there's a section on it in the Integration chapter


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    #3
    .:FuZion:.
    Closed Accounts Posts: 160 ✭✭.:FuZion:.


    Im trying to do it. Stuck as far as (4cos(Squared)Q)^1/2(2cosQ)dQ.... :rolleyes:


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    #4
    Scoobydooo
    Registered Users Posts: 341 ✭✭Scoobydooo


    Ye i think its from the 1999 papers any help is greatly appreciated


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    #5
    Des23
    Closed Accounts Posts: 287 ✭✭Des23


    Fill in 2sinQ into x^2 and you also have to fill it into dx, so you have to differentiate to get 2cosQdQ (change the limits) after that just do all the trig and you have to integrate 2(1 + cos2Q) between pi/3 and 0


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    #6
    .:FuZion:.
    Closed Accounts Posts: 160 ✭✭.:FuZion:.


    Forgot to change the limits. :rolleyes: :p


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    #7
    Challenged
    Closed Accounts Posts: 40 ✭Challenged


    http://studentxpress.ie/papers/intsoln1999.pdf


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    #8
    Scoobydooo
    Registered Users Posts: 341 ✭✭Scoobydooo


    Thanks for all your help guys !! :)


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    #9
    .:FuZion:.
    Closed Accounts Posts: 160 ✭✭.:FuZion:.


    Did you get it out? :pac: Im doing 2004 now. :p


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