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    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
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    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    dx/d(theta) = ?
    Then multiply by this.
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    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    Hint:

    x=\sin \theta \Rightarrow \dfrac{dx}{d\theta} = \cos \theta \Rightarrow dx = \cos \theta \ d\theta

    You don't just substitute sin x .

    Note that you also have to change the limits.
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    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    Before the 'next':
    After substitution you should to get \frac{1}{\cos^3 \theta} for the integrandus.

    And what will you get for the integrating factor of dx with substitution
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    Can some one please post a full solution so I can work through it...
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    Im stuck myself I got to this 1/(cos^2t)^3/2dx
    1/(cos^2t)^3/2 (1/cost) dx
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    (Original post by IShouldBeRevising_)
    Can some one please post a full solution so I can work through it...
    Try what people have suggested first. Cannot give full solutions here.
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    This A-level maths is funny compared to IB HL...

    Right so you got 1/(cost)^3 dx ... you also know that dx=cost dt
    giving 1/cos²t dt

    you also need to find the new bounds since x=sint
    giving lower bound 0 and upper bound pi/4

    So youre left to find the definite integral between 0 and pi/4 of 1/cos²t dt...

    Which integrates to tant. now you plug in pi/4 and 0. and subtract both results.

    tan(pi/4)-tan(0) = 1-0 = 1


    If you don't understand a step, feel free to quote and ask.
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    (Original post by IShouldBeRevising_)
    Can some one please post a full solution so I can work through it...
    Why don't you follow the suggestions given and post your working and then we can comment.

    You were nearly there at the start, except you put in sin x where you should have put sin \theta. You can then simplify the denominator, and you also need to convert dx into d\theta using dx/d\theta
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    (Original post by Giveme45)
    This A-level maths is funny compared to IB HL...

    Right so you got 1/(cost)^3 dx ... you also know that dx=cost dt
    giving 1/cos²t dt

    you also need to find the new bounds since x=sint
    giving lower bound 0 and upper bound pi/4

    So youre left to find the definite integral between 0 and pi/4 of 1/cos²t dt...

    Which integrates to tant. now you plug in pi/4 and 0. and subtract both results.

    tan(pi/4)-tan(0) = 1-0 = 1


    If you don't understand a step, feel free to quote and ask.


    wrong, the limits you have are incorrect remember that x=sin theta so theta = arcsin x and hence you get the limits
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    (Original post by giobru)
    wrong, the limits you have are incorrect remember that x=sin theta so theta = arcsin x and hence you get the limits
    True, admitedly made a stupid mistake there, didn't pay enough attention, gj for pointing out
 
 
 
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