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Integrate 1 + sinx/cos x

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    I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
    The book also says let u=sin x

    Now I was thinking of changing it into sec x + tan x but that didn't get me far

    Should I bring cosx to the top?

    Sorry btw, I can't use latex so please bear with me!
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    If you substitute u=\sin x then you get dx = \dfrac{du}{\cos x}, so you get \cos^2 x on the denominator. How might you write \cos^2 x in terms of u?
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    (Original post by nuodai)
    If you substitute u=\cos x then you get dx = \dfrac{du}{\cos x}, so you get \cos^2 x on the denominator. How might you write \cos^2 x in terms of u?
    U^2

    e: i got negged coz i got it wrong :facepalm2:
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    So I got 1 + sinx du/cos^2x?
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    (Original post by This Honest)
    U^2
    No. u=\sin x so u^2 = \sin^2 x \ne \cos^2 x - try again!

    (Original post by This Honest)
    So I got 1 + sinx du/cos^2x?
    Not quite, you should have \dfrac{1 + \sin x}{\cos^2 x} du. Now put this in terms of u.
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    (Original post by This Honest)
    So I got 1 + sinx du/cos^2x?
    Then use the fact that u= sinx

    and cos^2x = 1-sin^2x.
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    (Original post by nuodai)
    No. u=\sin x so u^2 = \sin^2 x \ne \cos^2 x - try again!



    Not quite, you should have \dfrac{1 + \sin x}{\cos^2 x} du. Now put this in terms of u.

    (Original post by f1mad)
    Then use the fact that u= sinx

    and cos^2x = 1-sin^2x.
    Alright guys,

    I've now got: 1 + u/1-sin^x
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    (Original post by This Honest)
    Alright guys,

    I've now got: 1 + u/1-sin^x
    Write it in terms of u; that still has an x in it. You'll notice that you can factorize something that simplifies the expression into something you can integrate.
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    (Original post by This Honest)
    I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
    The book also says let u=sin x

    Now I was thinking of changing it into sec x + tan x but that didn't get me far

    Should I bring cosx to the top?

    Sorry btw, I can't use latex so please bear with me!
    This question can also be easily done without substitution,

     \displaystyle \int \left(1+\frac{sinx}{cosx}\right) dx

    Now try to differentiate,  \displaystyle u=ln(cosx)
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    integrate: 1 +u/1-u^2 ??
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    (Original post by This Honest)
    I'm quite stuck on this question: Integrate 1 + sinx/cosx by substitution.
    The book also says let u=sin x

    Now I was thinking of changing it into sec x + tan x but that didn't get me far

    Should I bring cosx to the top?

    Sorry btw, I can't use latex so please bear with me!
    I'm sorry but i'm too busy to help you
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    (Original post by This Honest)
    Alright guys,

    I've now got: 1 + u/1-sin^x
    1+u/1-u^2

    remember that u = sinx and that cos^2(x) = 1 - sin^2(x)
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    (Original post by nuodai)
    Write it in terms of u; that still has an x in it. You'll notice that you can factorize something that simplifies the expression into something you can integrate.
    :woo: 1/1-u
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    (Original post by This Honest)
    integrate: 1 +u/1-u^2 ??
    Yes, remember you will have to use partial fraction.

     \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}
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    I GOT IT!

    -ln!1-sinx! + c

    can't find modular signs on keyboard $

    Thanks everyone for the help
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    (Original post by This Honest)
    integrate: 1 +u/1-u^2 ??
    have you learnt the integration rule, where if you differentiate the denominator, you get the value that appears in the numerator. hence ln(your original function)

    have you come across this before?

    this could prove useful here.

    someone else correct me if im wrong

    WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:
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    (Original post by raheem94)
    Yes, remember you will have to use partial fraction.

     \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}
    I didn't quite understand your method but thanks anyways
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    (Original post by blacklistmember)
    I'm sorry but i'm too busy to help you
    Why post then?
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    (Original post by James A)
    have you learnt the integration rule, where if you differentiate the denominator, you get the value that appears in the numerator. hence ln(your original function)

    have you come across this before?

    this could prove useful here.

    someone else correct me if im wrong

    WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:
    OHHHHHHH YEAHHHHHHH!

    The book said "by sub" so my mind was only on sub and nothing else
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    (Original post by raheem94)
    Yes, remember you will have to use partial fraction.

     \displaystyle 1 + \frac{u}{1-u^2} = 1 + \frac{u}{(1-u)(1+u)}
    I don't think so.

    If it's (1+u)/ (1-u^2) = (1+u)/(1-u)(1+u) = 1/(1-u) = - I of -1/(1-u) then it's a simple f'(u)/f(u) integral.

    I think you mis-interpreted .

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