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    (Original post by Proflash)
    lol that question is on Q3 page 140 not 142
    i got the older version of the book, so there is only 1 page on exam style questions and i guess it was a coincidence that question 3 was substitution too, so i automatically thought she was asking about that question.
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    (Original post by nukethemaly)
    Oh yeah and I've already done that one. It's just the trig substitutions I have trouble with.
    what was the answer?
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    (Original post by nukethemaly)
    Yeah that one! I've attached my neatest try! and also the question

    EDIT: I literally don't know what i'm trying to do with those limits there haha
    That's a nice question, plenty of trigonometry in it.

    OK, firstly, you got \dfrac{dx}{d\theta}. There's no need to write it as \dfrac{d\theta}{dx} in this case; just substitute in:
    \sec \theta \tan \theta d\theta
    \sec^2 \theta
    and don't forget to change the limits.

    So:

    \displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

    Write it as:

    \displaystyle \int_\frac{\pi}{3}^{arccos(\frac  {1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

    Now, how else can you write sec^2 \theta - 1 ? Then do some simplifying and you should get it.
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    (Original post by masryboy94)
    what was the answer?
    I'm getting 0.0940 (to 4 s.f).
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    (Original post by usycool1)
    That's a nice question, plenty of trigonometry in it.

    OK, firstly, you got \dfrac{dx}{d\theta}. There's no need to write it as \dfrac{d\theta}{dx} in this case; just substitute in \sec \theta \tan \theta d\theta, \sec^2 \theta and don't forget to change the limits.

    So:

    \displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

    Write it as:

    \displaystyle \int_\frac{\pi}{3}^{arccos(\frac  {1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

    Now, how else can you write sec^2 \theta - 1 ? Then do some simplifying and you should get it.
    (Original post by usycool1)
    I'm getting 0.0940 (to 4 s.f).
    haha you beat me to it, yep got that answer ! just like you said hope you get it now Aly
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    (Original post by usycool1)
    I'm getting 0.0940 (to 4 s.f).
    out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.
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    (Original post by usycool1)
    That's a nice question, plenty of trigonometry in it.

    OK, firstly, you got \dfrac{dx}{d\theta}. There's no need to write it as \dfrac{d\theta}{dx} in this case; just substitute in:
    \sec \theta \tan \theta d\theta
    \sec^2 \theta
    and don't forget to change the limits.

    So:

    \displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

    Write it as:

    \displaystyle \int_\frac{\pi}{3}^{arccos(\frac  {1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

    Now, how else can you write sec^2 \theta - 1 ? Then do some simplifying and you should get it.
    holy ****, it's that easy!
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    (Original post by masryboy94)
    haha you beat me to it, yep got that answer ! just like you said hope you get it now Aly
    Yep I do!

    Thanks you guys!
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    (Original post by masryboy94)
    out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.
    I haven't looked at what he's done, but I'd simplify like this:
    \dfrac{sec(x)tan(x)}{tan^3(x)}=\dfrac{sec(x)tan(x)}{tan(x)(tan^  2(x))}=\dfrac{sec(x)}{sec^2(x)-1}
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    (Original post by reubenkinara)
    I haven't looked at what he's done, but I'd simplify like this:
    \dfrac{sec(x)tan(x)}{tan^3(x)}=\dfrac{sec(x)tan(x)}{tan(x)(tan^  2(x))}=\dfrac{sec(x)}{sec^2(x)-1}
    but then how would you integrate that?
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    (Original post by masryboy94)
    but then how would you integrate that?
    Jut realized the context. :eek:
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    i've just been doing a soloman paper now and on the mark scheme to a bionomial question they gave the answer as follows: (attachment). my problem is i got the coefficient of x3 to be 160 however they got -160. but if you were to look at their calculation, its should be (4x8) - (2 x -64) which is 160. so have they done a mistake there with the sign? or am i not catching on something?
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    (Original post by masryboy94)
    but then how would you integrate that?

    i tried to sub in x=sec and attempted the f'(x)/f(x) integration but im getting it wrong for some reason.
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    (Original post by masryboy94)
    i've just been doing a soloman paper now and on the mark scheme to a bionomial question they gave the answer as follows: (attachment). my problem is i got the coefficient of x3 to be 160 however they got -160. but if you were to look at their calculation, its should be (4x8) - (2 x -64) which is 160. so have they done a mistake there with the sign? or am i not catching on something?
    I did that paper today. The mark-scheme is wrong.
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    Rather un-intuitive but use the substitution u=tanx
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    (Original post by masryboy94)
    but then how would you integrate that?
    Wouldn't it be -(sec^2 (x)-1)^(-0.5) if you diffentiate that you should get that function.
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    Hey!

    I'm stuck on question 3 and 8 in the C4 specimen paper. I'm in the early stages of solving and would really appreciate if any solves them for me as they are not in ExamSolutions.

    Thank you in advance!
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    (Original post by Vaner)
    Hey!

    I'm stuck on question 3 and 8 in the C4 specimen paper. I'm in the early stages of solving and would really appreciate if any solves them for me as they are not in ExamSolutions.

    Thank you in advance!
    I'm going out to the movies right now (Iron Man 3 :cool:), but if no one's helped you out when I get back, I'll take a look at it.
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    (Original post by Vaner)
    Hey!

    I'm stuck on question 3 and 8 in the C4 specimen paper. I'm in the early stages of solving and would really appreciate if any solves them for me as they are not in ExamSolutions.

    Thank you in advance!
    The integration by substitution and differential equation question? What have you tried?

    (Original post by masryboy94)
    out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.
    I simplified it down to -cosec x

    (Original post by Lanre)
    Wouldn't it be -(sec^2 (x)-1)^(-0.5) if you diffentiate that you should get that function.
    Nope, try differentiating it.
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    (Original post by usycool1)



    I simplified it down to -cosec x

    ooo so you got to the point cosx/sin^2 x and then brought the sin^2 x to the top, did the reverse chain rule to get -1/sinx hence getting the -cosecx, right?
 
 
 
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