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    (Original post by chubby)
    thanks a lot fermat!!!! i learnt something new today. thanks
    could u also do 2 (iii). i know it is a show that question but i got the end answer as (11/18) instead of (26/9). pls help if u can.
    int 7sinhx-sinh2x dx
    =1/2int 7 {e^x-e^(-x)}-e^(2x)+e^(-2x)
    =1/2 {[7(e^x+e^(-x))]-1/2(e^(2x)+e^(-2x))}
    x=0 gives 1/2{7+7-1/2(2)]=13/2
    x=ln 3
    e^x=e^(ln3)=3
    e^(-x)=e^-ln3=e^ln(1/3)=1/3
    e^(2x)=e^(2ln3)=e^ln9=9
    e^(-2x)=e^-2ln3=e^ln(1/9)=1/9
    so have 1/2{7(3+1/3)-1/2(9+1/9)}=1/2(70/3-41/9)=169/18
    hence int=
    169/18-(13/2)=52/18
    =26/9
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    thanks. did it. made a very silly mistake. forgot to mulitiply by 0.5 in the second term. thanks anyway
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    Oooh, I actually did that paper (questions 1, 2, 3). It wasn't nice.
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    (Original post by chubby)
    Also fermat in 4(i) the curve is between 0 and pi as mentioned in the question so i am guessing all the four loops lie above the xaxis (two between 0 and pi/2 and 2 between pi/2 and pi). the second and fourth loop starting from the 0 angled line are dotted lines. am i right?
    Hi chubby,
    these things can sometimes be a bit confusing.
    The curve doesn't lie between 0 and pi.
    That is the range of the argument , theta.

    Between 0 and pi/4, sin4θ is +ve so r is +ve and the 1st loop is gotten.
    Between pi/4 and pi/2, sin4θ is -ve, so r is -ve and the curve now appears as a 2nd loop on the other side of the origin - denoted as a broken line.
    Between θ/2 and 3pi/4, sin4θ is +ve again, r is +ve, and the 3rd loop is gotten.
    Between 3pi/4 and pi, sin4θ is -ve again, r is -ve and the curve again appears as a 4th loop on the other side of the origin - denoted as a broken line.

    So, the curve appears in all quadrants with the 1st and 3rd loops above the x-axis and the 2nd and 4th loops below the x-axis.

    The Fig attached shows how the loops would look if you were taking r as +ve all the time. eg. r = k|sinθ|.
    Attached Images
     
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    YAY. thanks fermat!!!! get it now.
 
 
 
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