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P3 Integration (god I hate P3) watch

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    Anyone want to suggest a method for integrating....

    2tanx / cos^2x
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    1/cos^2x is the same as sec^2x
    so the integral is in the form f(x)f'(x) and you can integrate it directly
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    2tanx/cos^2x is 2sinx/cos^3 x
    solution is cos^-2 x
    just guess.
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    (Original post by Leekey)
    Anyone want to suggest a method for integrating....

    2tanx / cos^2x
    Rewrite tanx as sinx/cosx substitution of u = cos x, du/dx = -sinx, dx = -du/sinx
    int 2sinx/cos^3 x dx = int -2/u^3 du = int -2u^(-3) du = u^(-2) + c = sec^2 x + c
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    (Original post by Bezza)
    Rewrite tanx as sinx/cosx substitution of u = cos x, du/dx = -sinx, dx = -du/sinx
    int 2sinx/cos^3 x dx = int -2/u^3 du = int -2u^(-3) du = u^(-2) + c = sec^2 x + c
    too complicated....
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    (Original post by minkailin)
    too complicated....
    I'd prefer to call it working through the answer. Not everyone is going to be able to guess it straight away
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    I think the best method is the one that p8224 suggested, i.e. to re-write the 1/(cos x)^2 as (sec x)^2, which now makes the integration easy
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    (Original post by p8224)
    1/cos^2x is the same as sec^2x
    so the integral is in the form f(x)f'(x) and you can integrate it directly
    I've stared at the thing for 5 minutes and I didn't spot that!!! Thankyou (rep on way!!!)!!!
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    wow...
    there are 2 answers!
    (tanx)^2 + c and
    (secx)^2 + k

    because (tanx)^2 + 1 (a constant) = (secx)^2......just thought it's nice to know
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    (Original post by mockel)
    I think the best method is the one that p8224 suggested, i.e. to re-write the 1/(cos x)^2 as (sec x)^2, which now makes the integration easy
    His way you'll end up with tan^2 x though - sec^2 x is much more fun to say!!
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    (Original post by Leekey)
    Thankyou (rep on way!!!)!!!
    aww...!! thank you!
    and you're most welcome! :cool:
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    [QUOTE=Bezza]His way ....[QUOTE]

    I'm a 'her' btw!
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    (Original post by p8224)
    I'm a 'her' btw!
    Sorry! It just seems that most people answering maths questions are 'hims'
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    (Original post by Bezza)
    Sorry! It just seems that most people answering maths questions are 'hims'
    hahahhaha!! its alright!
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    my attachment answers the question i think. Well it makes sense for me anyway
    sorry i cant do the integral sign

    Edited:
    sorry i have made an error in my integration,coz I left out the "2".

    so the answer is, if u just take out the 2 from the integral as a multiplier, and then multiply the 1/2 by 2, ud end up with (cosx)^-2 = (secx)^2

    so sorry about the error, but its hard to do on the computer correctly
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    (Original post by p8224)
    wow...
    there are 2 answers!
    (tanx)^2 + c and
    (secx)^2 + k

    because (tanx)^2 + 1 (a constant) = (secx)^2......just thought it's nice to know
    no tanx squared is incorrect.
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    (Original post by ResidentEvil)
    no tanx squared is incorrect.
    it is correct, because there is an alternate method to approach this question.
    For example, 1/cos²x = sec²x

    so the question can be altered to look like this:

    2[INT (tanx)(sec²x) dx]

    This is now in the form INT f(x).f '(x) dx

    so the integration can be done straight away:

    = 2 [ (1/2)(tanx)²]
    = tan²x + c
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    Both answers are correct,because:
    (tanx)^2 + C =(secx)^2 -1 + C =(secx)^2 + D. (let C-1=D)
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    I see yer point now, but I personally dont like that approach
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    (Original post by ResidentEvil)
    I see yer point now, but I personally dont like that approach
    i guess everyone has there own approaches that they like.
 
 
 
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