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    (Original post by MathsNerd1)
    It shouldn't be that bad, but who doesn't love Maths Lets see how you do with integration and it can be done using C4 techniques too

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    I = \displaystyle\int \dfrac{4\tan^2(x)}{1-\tan^4(x)} \ dx



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    that is a nice question, i can only see partial fractions in this?
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    (Original post by masryboy94)
    that is a nice question, i can only see partial fractions in this?
    Indeed And I wouldn't go down the partial fraction route, instead I separated it all into sin's and cos's, also remember your identities and double angle formulas and you should slowly get there I think this one took me about 10 minutes to do, trying two different approaches too
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    (Original post by MathsNerd1)
    It shouldn't be that bad, but who doesn't love Maths Lets see how you do with integration and it can be done using C4 techniques too

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    I = \displaystyle\int \dfrac{4\tan^2(x)}{1-\tan^4(x)} \ dx



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    i got it to be

     8tanxsec^2x its probably soooo wrong ! lool
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    (Original post by masryboy94)
    ONLY if you have time, could you pick out the hardest questions in the soloman papers for C4 and go through them?
    Sure
    I will have a look next week once the GCSE 100 videos are sorted!
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    (Original post by masryboy94)
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    i got it to be

     8tanxsec^2x its probably soooo wrong ! lool
    Could you show your working as that is quite a way off the actual answer. What have you actually done and to get that?
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    (Original post by m4ths/maths247)
    Sure
    I will have a look next week once the GCSE 100 videos are sorted!
    thank you ! i know you must be very busy so me and probably everyone else appreciates what you are doing, again thank you very much for your help, good luck with your class and the videos

    p.s. could you let us know when you put up the video on this forum? thank you
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    (Original post by masryboy94)
    thank you ! i know you must be very busy so me and probably everyone else appreciates what you are doing, again thank you very much for your help, good luck with your class and the videos

    p.s. could you let us know when you put up the video on this forum? thank you
    Sure, will do
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    (Original post by MathsNerd1)
    Could you show your working as that is quite a way off the actual answer. What have you actually done and to get that?
    haha dw i expected that, i was improvising much of the way haha and i realised i did VITAL mistakes on the way haha. i was just seeing if partial fractions work but it probably just complicates everything
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    (Original post by MathsNerd1)
    Indeed And I wouldn't go down the partial fraction route, instead I separated it all into sin's and cos's, also remember your identities and double angle formulas and you should slowly get there I think this one took me about 10 minutes to do, trying two different approaches too
    hey i had a quick go at it and i got (not sure if this right) :

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    \frac{1}{2}sin2x+x+C
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    (Original post by masryboy94)
    haha dw i expected that, i was improvising much of the way haha and i realised i did VITAL mistakes on the way haha. i was just seeing if partial fractions work but it probably just complicates everything
    Fair enough then and I wouldn't suggest a substitution as that would also over complicate the question, I'd just try and simplify the expression as much as possible and then use a certain technique to get the final answer
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    (Original post by nm786)
    hey i had a quick go at it and i got (not sure if this right) :

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    \frac{1}{2}sin2x+x+C
    That's unfortunately incorrect :-/ Did you try to simplify the integral as much as you possibly could before integrating? Remember you double angle formulas and trig identities might come in use for this question
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    (Original post by MathsNerd1)
    That's unfortunately incorrect :-/ Did you try to simplify the integral as much as you possibly could before integrating? Remember you double angle formulas and trig identities might come in use for this question
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    i got to  \frac {sin^22x}{cos^4x -sin^4x} is that right? or am i going the wrong way
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    (Original post by MathsNerd1)
    That's unfortunately incorrect :-/ Did you try to simplify the integral as much as you possibly could before integrating? Remember you double angle formulas and trig identities might come in use for this question
    this is what i did:
    doesn't the integral simplify to: \frac{4}{1 - \frac{sin^2x}{cos^2x}}? and then i used the identity: 1+tan^2x=sec^2x and ended up with \frac{2}{-sec^2x} ---> -2cos^2x and went on from here :/
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    Here is the diagram sorry:


    and question:

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    Name:  Picture1.jpg
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    how did they figure out the value for t?
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    (Original post by masryboy94)
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    i got to  \frac {sin^22x}{cos^4x -sin^4x} is that right? or am i going the wrong way
    Well the top line is correct but the bottom one can be simplified to give a simplified integral then you need to use a certain technique to integrate the integral
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    (Original post by nm786)
    this is what i did:
    doesn't the integral simplify to: \frac{4}{1 - \frac{sin^2x}{cos^2x}}? and then i used the identity: 1+tan^2x=sec^2x and ended up with \frac{2}{-sec^2x} :/
    Hmm I'm not quite sure how you got it to simplify into that :-/ What can the denominator be expressed as? Then is there something you know that can simplify it? Once you've done that is there anyway you could simplify the fraction? Let me know if you're still stuck

    Edit: It looks like you're missing a function in your numerator, other than that's its all going the correct way so far
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    (Original post by MathsNerd1)
    Well the top line is correct but the bottom one can be simplified to give a simplified integral then you need to use a certain technique to integrate the integral
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    well i got to  \frac{sin^22x}{cos2x} and now i've just gone blank lool

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    (Original post by masryboy94)
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    well i got to  \frac{sin^22x}{cos2x} and now i've just gone blank lool
    Okay so how else can that be written? Then maybe try parts on it to see if you can get the answer
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    (Original post by MathsNerd1)
    Okay so how else can that be written? Then maybe try parts on it to see if you can get the answer
    for top and bottom?
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    (Original post by masryboy94)
    for top and bottom?
    Yeah, I'm sure you can rewrite what you've got as a combination of trigs? Then all you'll have to do is IBP and you'll get the correct answer
 
 
 
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