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    I know that this is only a single question so sorry about making a new thread for it...but is the integral of artanhx

    x.artanhx+½ln l1-x^2l ?

    I got this by parts but a friend got something different and I was hoping someone could tell me whether than answer is right :/

    THanks

    Also, does anybody know where mark schemes for soloman papers G and H could be found?

    If not is anybody able to do Paper G question 3?

    Find ∫ 1/(1+sin2x) dx

    between 0 and pi/4....I used the t substitution where t=tanx, since I was dealing with sin2x and wondered whether this would be ok...although I take it that dy/dt would actually be 1/(1+t^2)? I got 0.5 anyway...but if someone could tell me the correct answer, I would be grateful. Thanks
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    PM me ur email address and ill email you the markschemes.
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    (Original post by lgs98jonee)
    I know that this is only a single question so sorry about making a new thread for it...but is the integral of artanhx

    x.artanhx+½ln l1-x^2l ?
    Yes, although since arctanh is only defined on (-1, 1) it is pointless to put the abs signs in.
    Find \int 1/(1+sin2x) dx between 0 and pi/4
    Using cos(2x) = 2cos^2(x) - 1,

    \int_0^{\pi/4}1/(1 + sin(2x)) dx
    = \int_0^{\pi/4}1/(1 + cos(2x)) dx
    = \int_0^{\pi/4}(1/2)sec^2(x) dx
    = [(1/2)tan(x)](from 0 to pi/4)
    = 1/2
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    (Original post by Jonny W)
    Yes, although since arctanh is only defined on (-1, 1) it is pointless to put the abs signs in.
    oh ok...thanks alot :-)
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    (Original post by lgs98jonee)
    I know that this is only a single question so sorry about making a new thread for it...but is the integral of artanhx

    x.artanhx+½ln l1-x^2l ?

    I got this by parts but a friend got something different and I was hoping someone could tell me whether than answer is right :/

    THanks

    ...
    A way of checking this is to put both expressions into a graphical calculator or graph-plotting program.
    If both expressions give the same curve then they are equivalent and are correct (but alternative) solutions for the integral of artanhx.

    If the curves don't match up then you gotta do a little bit more work to find out which one is wrong.
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    (Original post by Jonny W)
    Yes, although since arctanh is only defined on (-1, 1) it is pointless to put the abs signs in.

    Using cos(2x) = 2cos^2(x) - 1,

    \int_0^{\pi/4}1/(1 + sin(2x)) dx
    = \int_0^{\pi/4}1/(1 + cos(2x)) dx
    = \int_0^{\pi/4}(1/2)sec^2(x) dx
    = [(1/2)tan(x)](from 0 to pi/4)
    = 1/2
    eh...what have you done....you just swictehd from sin to cos...how?
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    (Original post by lgs98jonee)
    eh...what have you done....you just swictehd from sin to cos...how?
    If you think of the graphs of sin2x between 0 and pi/4 and cos2x between 0 and pi/4 then you'll see the are mirror images of one another in the x = pi/8 line so all the above is valid area-wise

    If you wish to think of it substitution-wise though then set u = pi/4 -x and you'll see that the integrals are equal
 
 
 
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