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    In part (i) of this question it is established that if  y = \ln(1 + \sin x) - \ln(\cos x)

then  \frac{{\rm d}y}{{\rm d}x}=\frac{1}{\cos x}

    In part (ii) you use this to carry out \int_0^{\frac{1}{3}\pi}\sec \frac{1}{3}x\;{\rm d}x

    Now due to the  \frac{1}{3}x I get a multiplier of 3 which I carry through to a final =3\left[\ln{(2+\sqrt 3)} \right]

    The Mark scheme has no multiplier of 3 and carries no detail it just says
    Indefinite Integral =\left[\ln{(2+\sqrt 3)} \right]

    Here is my working (well the first line which carries all the way through)  

=3\left[ \ln \left(1+\sin x\right)-\ln \left(\cos x\right)\right]_0^\frac{\pi}{3}




    Where have I gone wrong?
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    (Original post by nerak99)
    The Mark scheme has no multiplier of 3 and carries no detail it just says
    Indefinite Integral =\left[\ln{(2+\sqrt 3)} \right]

    Where have I gone wrong?
    Paper I looked at looks for the integral of sec x, not sec x/3.

    Assuming you'e not misread the question, I suspect a revision in the paper, and there's a mismatch between the markscheme you're looking at and the paper you're looking at.
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    Aah thanks, I was picking the question up from a reprinted version! Duh
 
 
 
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