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    (Original post by Converse girl)
    hi guys can you help me out iam having trouble intergratiting sec^2xtanx
    You could do it by substitution or, more quickly, by recognition. There are two possible answers for this, btw, there's just a different constant for both answers.
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    (Original post by justinawe)
    use the substitution u= \sec^2 x

    (Original post by brittanna)
    This is in the form f'(x)f(x). Try considering tan^2(x).

    (Original post by usycool1)
    You could do it by substitution or, more quickly, by recognition. There are two possible answers for this, btw, there's just a different constant for both answers.
    i got sec^2x
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    (Original post by Converse girl)
    i got sec^2x
    +C

    And you should have got (1/2)sec^2 x + C...what did you do to get that answer?
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    (Original post by Converse girl)
    i got sec^2x
    Close, but you forgot a part of it

    remember \displaystyle \int x^n \mathrm{ \ } dx = \frac{x^{n+1}}{n+1} + C
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    (Original post by Converse girl)
    i got sec^2x
    I got tan^2x/2 + c I used u=tanx someone confirm please
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    (Original post by QwertyG)
    I got tan^2x/2 + c I used u=tanx someone confirm please
    That's right too, just with a different constant. :yy:
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    (Original post by usycool1)
    +C

    And you should have got (1/2)sec^2 x + C...what did you do to get that answer?

    (Original post by justinawe)
    Close, but you forgot a part of it

    remember \displaystyle \int x^n \mathrm{ \ } dx = \frac{x^{n+1}}{n+1} + C
    oh yeah i get it always forget +C

    thanks guys
    i need a lot of practice with intergration :stomp:
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    (Original post by usycool1)
    That's right too, just with a different constant. :yy:
    If you done it with substitution would you have used u=tanx or u=sec^2x?
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    (Original post by QwertyG)
    If you done it with substitution would you have used u=tanx or u=sec^2x?
    I would have used u= \tan x looking at the question properly. I said u = \sec^2 x earlier but wasn't really thinking, the former would have been much easier.
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    (Original post by usycool1)
    +C

    And you should have got (1/2)sec^2 x + C...what did you do to get that answer?
    I see what you mean now! I used integration by parts 3 times and still continues
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    (Original post by otrivine)
    I see what you mean now! I used integration by parts 3 times and still continues
    By parts 3 times?!! :eek:
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    (Original post by otrivine)
    I see what you mean now! I used integration by parts 3 times and still continues
    Here's how you can do it (solution in spoiler):

    Spoiler:
    Show
    \displaystyle\int e^{3x} cos 2x \ dx

    Use integration by parts. Let "u" equal \cos 2x and \frac{dv}{dx}=e^{3x}

    So...

    \frac{du}{dx}=-2 \sin 2x and v = \frac{1}{3}e^{3x}

    So you should get (after using integration by parts):

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x - \displaystyle\int {-\dfrac{2}{3}e^{3x} \sin 2x} dx

    Which is the same as:

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3} \displaystyle\int e^{3x} \sin 2x dx

    Now use integration by parts again to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3}(\dfrac{1}{3}e^{3x} \sin 2x - \dfrac{2}{3}\displaystyle\int e^{3x} \cos 2x dx)

    Expand the brackets to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x - \dfrac{4}{9} \displaystyle\int e^{3x} \cos 2x dx

    OK, now we'll go on forever and ever if we keep using integration by parts. So:

    To make things a tiny bit easier, let I = \displaystyle\int e^{3x} cos 2x dx

    So we have:

    I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x -\dfrac{4}{9}I

    Take the "I" terms onto one side:

    I + \dfrac{4}{9}I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    Now "solve" for I:

    I(1+\dfrac{4}{9})=\dfrac{1}{3}e^  {3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    \dfrac{13}{9}I=\dfrac{1}{3}e^{3x  } \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    So after "dividing" both sides by \dfrac{13}{9}, you should eventually (after simplifying a bit) get:

    I=\dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)

    \displaystyle\int e^{3x} cos(2x) = \dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)+C
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    (Original post by usycool1)
    Here's how you can do it (solution in spoiler):

    Spoiler:
    Show
    \displaystyle\int e^{3x} cos 2x \ dx

    Use integration by parts. Let "u" equal \cos 2x and \frac{dv}{dx}=e^{3x}

    So...

    \frac{du}{dx}=-2 \sin 2x and v = \frac{1}{3}e^{3x}

    So you should get (after using integration by parts):

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x - \displaystyle\int {-\dfrac{2}{3}e^{3x} \sin 2x} dx

    Which is the same as:

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3} \displaystyle\int e^{3x} \sin 2x dx

    Now use integration by parts again to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3}(\dfrac{1}{3}e^{3x} \sin 2x - \dfrac{2}{3}\displaystyle\int e^{3x} \cos 2x dx)

    Expand the brackets to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x - \dfrac{4}{9} \displaystyle\int e^{3x} \cos 2x dx

    OK, now we'll go on forever and ever if we keep using integration by parts. So:

    To make things a tiny bit easier, let I = \displaystyle\int e^{3x} cos 2x dx

    So we have:

    I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x -\dfrac{4}{9}I

    Take the "I" terms onto one side:

    I + \dfrac{4}{9}I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    Now "solve" for I:

    I(1+\dfrac{4}{9})=\dfrac{1}{3}e^  {3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    \dfrac{13}{9}I=\dfrac{1}{3}e^{3x  } \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    So after "dividing" both sides by \dfrac{13}{9}, you should eventually (after simplifying a bit) get:

    I=\dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)

    \displaystyle\int e^{3x} cos(2x) = \dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)+C
    I lost what was going on after line 2

    Oh wait never mind i get it
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    how far is evryone through c3 and c4 revision?
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    (Original post by usycool1)
    Here's how you can do it (solution in spoiler):

    Spoiler:
    Show
    \displaystyle\int e^{3x} cos 2x \ dx

    Use integration by parts. Let "u" equal \cos 2x and \frac{dv}{dx}=e^{3x}

    So...

    \frac{du}{dx}=-2 \sin 2x and v = \frac{1}{3}e^{3x}

    So you should get (after using integration by parts):

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x - \displaystyle\int {-\dfrac{2}{3}e^{3x} \sin 2x} dx

    Which is the same as:

    \displaystyle\int e^{3x} cos 2x \ dx = \frac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3} \displaystyle\int e^{3x} \sin 2x dx

    Now use integration by parts again to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{3}(\dfrac{1}{3}e^{3x} \sin 2x - \dfrac{2}{3}\displaystyle\int e^{3x} \cos 2x dx)

    Expand the brackets to get:

    \displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x - \dfrac{4}{9} \displaystyle\int e^{3x} \cos 2x dx

    OK, now we'll go on forever and ever if we keep using integration by parts. So:

    To make things a tiny bit easier, let I = \displaystyle\int e^{3x} cos 2x dx

    So we have:

    I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x -\dfrac{4}{9}I

    Take the "I" terms onto one side:

    I + \dfrac{4}{9}I = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    Now "solve" for I:

    I(1+\dfrac{4}{9})=\dfrac{1}{3}e^  {3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    \dfrac{13}{9}I=\dfrac{1}{3}e^{3x  } \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x

    So after "dividing" both sides by \dfrac{13}{9}, you should eventually (after simplifying a bit) get:

    I=\dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)

    \displaystyle\int e^{3x} cos(2x) = \dfrac{1}{13}e^{3x}(3 \cos 2x + 2 \sin 2x)+C

    wow! wait, can i now just use substitution, my integration by parts was the same as urs but do not get then how you or why you considered I?:confused:
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    (Original post by otrivine)
    wow! wait, can i now just use substitution, my integration by parts was the same as urs but do not get then how you or why you considered I?:confused:
    Its quite unlikely to come up if you ask me - I don't think it has ever before... where did you get the question from ?
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    (Original post by posthumus)
    Its quite unlikely to come up if you ask me - I don't think it has ever before... where did you get the question from ?
    Usycool1 asked this question and I was trying to solve it!

    By the way on page on differenitation example 4! I think they made a mistake ,can you or anyone confirm this.

    The question was

    find the valie of dy/dx at point (1,1)

    where 4xy2 +6x2/y = 10
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    (Original post by otrivine)
    Usycool1 asked this question and I was trying to solve it!

    By the way on page on differenitation example 4! I think they made a mistake ,can you or anyone confirm this.

    The question was

    find the valie of dy/dx at point (1,1)

    where 4xy2 +6x2/y = 10
    I checked it, looks right to me
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    (Original post by otrivine)
    Usycool1 asked this question and I was trying to solve it!

    By the way on page on differenitation example 4! I think they made a mistake ,can you or anyone confirm this.

    The question was

    find the valie of dy/dx at point (1,1)

    where 4xy2 +6x2/y = 10
    What answer did you get ?
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    (Original post by raiden95)
    I checked it, looks right to me
    they got 12x/y but should there not be a y with 12x?
 
 
 
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