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    Hi

    I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

    The integral of:

    cosx/(5-3sinx)

    Would it also be possible to tell me how you clever people do this question???

    There will probably be more questions later - but not for the moment!!!

    Thanks
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    (Original post by Silly Sally)
    Hi

    I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

    The integral of:

    cosx/(5-3sinx)

    Would it also be possible to tell me how you clever people do this question???

    There will probably be more questions later - but not for the moment!!!

    Thanks
    for this one use the substitution t=tan(x/2) (so get sinx and cosx in terms of t)
    by the way,what syallabus is this(its about p5 for edexcel)
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    (Original post by Silly Sally)
    Hi

    I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

    The integral of:

    cosx/(5-3sinx)

    Would it also be possible to tell me how you clever people do this question???

    There will probably be more questions later - but not for the moment!!!

    Thanks
    Well if you differentiate ln(5-3sinx) you'd get -3cosx/5-3sinx.

    So you want -1/3 cosx/5-3sinx + c.
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    (Original post by IntegralAnomaly)
    for this one use the substitution t=tan(x/2) (so get sinx and cosx in terms of t)
    by the way,what syallabus is this(its about p5 for edexcel)
    I am doing Edexcel - that can't be right!!! About P5 - this is from a P3 revision book- if you do Edexcel maths you may recognise the book - "revise for Pure mathematics 3" by Heinemann p28. I mean if you say it is P5 - i believe you - it is just really strange...
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    (Original post by theone)
    Well if you differentiate ln(5-3sinx) you'd get -3cosx/5-3sinx.

    So you want -1/3 cosx/5-3sinx + c.
    ohh crap,yes this method is the proper method,forget what i said(theone ur answer is not correct)
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    Do you do integration backwards - i.e. differentiate and see what you get and then find integral - cos i do that when it is simpler equation - but i can't do it with harder equations
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    (Original post by Silly Sally)
    Do you do integration backwards - i.e. differentiate and see what you get and then find integral - cos i do that when it is simpler equation - but i can't do it with harder equations
    use the substitution u=5-3sinx
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    (Original post by IntegralAnomaly)
    use the substitution u=5-3sinx
    Ahh - I understand substitution. Thanks for the tip.

    Btw - i would like to ask you all 2 questions rearding integration:

    1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

    2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

    Hope that makes sense???
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    (Original post by Silly Sally)
    Hi

    I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

    The integral of:

    cosx/(5-3sinx)

    Would it also be possible to tell me how you clever people do this question???

    There will probably be more questions later - but not for the moment!!!

    Thanks
    U=5-3SinX therefore U'=-3CosX which means dx = -du/3CosX

    Therefore CosX/(5-3SinX) -du/3CosX:

    = -1/3 * Integral of u^-1
    =-1/3lnU
    =-1/3ln(5-3SinX) + C

    Sorry, but I don't have the fancy shmancy keyboard signs.
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    (Original post by Silly Sally)
    Ahh - I understand substitution. Thanks for the tip.

    Btw - i would like to ask you all 2 questions rearding integration:

    1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

    2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

    Hope that makes sense???
    For part 2. Use the chain rule e.g:

    Y=(CosX)^4 say that U=CosX therefore Y=U^4

    dy/du = 4U^3

    You know that U=CosX therefore:

    du/dx = -SinX

    The chain rule says that dy/dx = dy/du * du/dx therefore:

    dy/dx = 4U^3 * -SinX and replace the U with CosX leaving:

    -4CosX^3SinX

    For part 1, its just practice.
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    (Original post by Bhaal85)
    For part 2. Use the chain rule e.g:

    Y=(CosX)^4 say that U=CosX therefore Y=U^4

    dy/du = 4U^3

    You know that U=CosX therefore:

    du/dx = -SinX

    The chain rule says that dy/dx = dy/du * du/dx therefore:

    dy/dx = 4U^3 * -SinX and replace the U with CosX leaving:

    -4CosX^3SinX
    .
    For the part 2 you were answering, i was referring to integration rathe than differentiation.
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    (Original post by Silly Sally)
    For the part 2 you were answering, i was referring to integration rathe than differentiation.
    Oh **** , sorry, what can I say, late night.
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    (Original post by Silly Sally)
    Hi

    I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

    The integral of:

    cosx/(5-3sinx)

    Would it also be possible to tell me how you clever people do this question???

    There will probably be more questions later - but not for the moment!!!

    Thanks
    u = 5-3sinx
    du = -3cosx dx

    integrate -3/u = -3lnu + c
    =-3ln(5-3sinx) + c

    is that right?
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    (Original post by Silly Sally)
    Ahh - I understand substitution. Thanks for the tip.

    Btw - i would like to ask you all 2 questions rearding integration:

    1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

    2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

    Hope that makes sense???
    u have the recognise(unlike i did above) the different forms of integration u can have,if u have something like f'(x)/f(x) then the integral will be ln|f(x)|+c,
    eg. ∫(sec²x/tanx)dx,u know that the derivative of tanx is sec²x,and this is of the form f'(x)/f(x) and so the answer is ln|tanx|+c.
    If u still are not confident with spotting it u can always use a substitution (u=f(x)).
    so have a go at this one:∫1/(xlnx)dx.
    If u have two completely different functions of x (x²e^x) then u should recognise that u have to use integration by parts.
    As for the trigonometric ones u have to try and get it in the form of f'(x)f(x).
    And then u can use a substitution u=f(x).
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    (Original post by Bhaal85)
    Oh **** , sorry, what can I say, late night.
    Its ok - it confused me a bit more than usual - and i am easily confused!!!

    So do you have a method for integration with high powers that i was referring to in a previous post? Would really appreciate it!!!
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    (Original post by imasillynarb)
    u = 5-3sinx
    du = -3cosx dx

    integrate -3/u = -3lnu + c
    =-3ln(5-3sinx) + c

    is that right?
    no.look again at the cosx term.
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    (Original post by Bhaal85)
    U=5-3SinX therefore U'=-3CosX which means dx = -du/3CosX

    Therefore CosX/(5-3SinX) -du/3CosX:

    = -1/3 * Integral of u^-1
    =-1/3lnU
    =-1/3(5-3SinX) + C

    Sorry, but I don't have the fancy shmancy keyboard signs.
    Typo:

    -1/3ln(5-3SinX) + C
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    (Original post by imasillynarb)
    u = 5-3sinx
    du = -3cosx dx

    integrate -3/u = -3lnu + c
    =-3ln(5-3sinx) + c

    is that right?
    Nearly - cos its integration - it is -1/3ln(5-3sinx)
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    (Original post by Silly Sally)
    2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

    with an odd power, e.g. sin^n x, you split it into sinx sin^(n-1) x, which becomes sinx - sinx cos^(n-1) x, which is a straightforward integral to cos^n x -cos x


    with an even power, you have to use the cos(2x) forumla

    e.g. sin²x = (1-cos2x)/2
    cos²x = (1+cos2x)/2

    so, eg. sin^2n x = (sin²x)^n = ((1-cos2x)/2)^n, and expand binomially
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    (Original post by Silly Sally)
    Its ok - it confused me a bit more than usual - and i am easily confused!!!

    So do you have a method for integration with high powers that i was referring to in a previous post? Would really appreciate it!!!
    I don't think that I have been taught how to do the integral of that trig:

    If I were doing it the method I would do, I would get:

    -CosX^6/6SinX + C which doesn't look right.
 
 
 

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