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    (Original post by James A)
    have you learnt the integration rule, where if you differentiate the denominator, you get the value that appears in the numerator. hence ln(your original function)

    have you come across this before?

    this could prove useful here.

    someone else correct me if im wrong

    WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:
    Look at post no. 9, i have hinted towards this method.
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    (Original post by blacklistmember)
    I'm sorry but i'm too busy to help you
    you wally.
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    (Original post by f1mad)
    Why post then?
    no idea
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    (Original post by James A)
    WHOOPS PARTIAL FRACTIONS ARE THE WAY FORWARD :woo:
    It's not.
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    (Original post by raheem94)
    Look at post no. 9, i have hinted towards this method.
    yeah i know, but i forgot to include your post in my response. Sorry.
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    (Original post by James A)
    you wally.
    i had to look that up :P
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    (Original post by f1mad)
    It's not.
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    (Original post by f1mad)
    It's not.
    Both ways, can be used. Your method is faster, but doing partial fractions also takes moments.
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    (Original post by This Honest)
    integrate: 1 +u/1-u^2 ??
    Please put in brackets with things like this, i.e. (1+u)/(1-u²). Not doing so has led to some misleading advice because people think you mean 1+\dfrac{u}{1-u^2} and 1 + \dfrac{\sin x}{\cos x} instead of \dfrac{1+u}{1-u^2} and \dfrac{1+\sin x}{\cos x}.

    Anyway, you got the right answer -\ln \left| 1 - \sin x \right| + C.

    (In fact you can remove the mod signs since 1-\sin x \ge 0 for all x.)
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    (Original post by raheem94)
    Both ways, can be used. Your method is faster, but doing partial fractions also takes moments.
    I beg to differ: your partial fraction isn't the same as the function of the substitution (see Nuodai's post).
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    (Original post by nuodai)
    Please put in brackets with things like this, i.e. (1+u)/(1-u²). Not doing so has led to some misleading advice because people think you mean 1+\dfrac{u}{1-u^2} and 1 + \dfrac{\sin x}{\cos x} instead of \dfrac{1+u}{1-u^2} and \dfrac{1+\sin x}{\cos x}.

    Anyway, you got the right answer -\ln \left| 1 - \sin x \right| + C.

    (In fact you can remove the mod signs since 1-\sin x \ge 0 for all x.)
    Sound like my maths teacher now
    Thanks, I will next time :yep:
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    (Original post by f1mad)
    I beg to differ: your partial fraction isn't the same as the function of the substitution (see Nuodai's post).
    So we were solving the wrong question. The question was different, right?
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    (Original post by This Honest)
    Sound like my maths teacher now
    Thanks, I will next time :yep:
    I've been made grumpy with age :p:
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    (Original post by raheem94)
    So we were solving the wrong question. The question was different, right?
    Yeah..sorry that was my fault. As noudai said...I caused confusion in the way I wrote it out
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    (Original post by nuodai)
    If you substitute u=\cos x then you get dx = \dfrac{du}{\cos x}, so you get \cos^2 x on the denominator. How might you write \cos^2 x in terms of u?
    Why have you made u equal to cos x here?
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    (Original post by This Honest)
    Yeah..sorry that was my fault. As noudai said...I caused confusion in the way I wrote it out
    No problem, but please learn LaTeX, it only takes a few minutes to learn it.
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    (Original post by raheem94)
    No problem, but please learn LaTeX, it only takes a few minutes to learn it.
    I'll try and learn it when I'm having difficulty with C4 again before I post the problem
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    (Original post by TheGrinningSkull)
    Why have you made u equal to cos x here?
    Typo; I'd set u=\sin x. If I'd actually set u=\cos x then I wouldn't have got dx = \dfrac{du}{\cos x}, after all.
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    An alternative approach would be to notice that \dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}, which after a bit of cancelling leads to logs directly.

    EDIT: just realised the OP was asked for a specific sub, so this can be ignored.
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    (Original post by Farhan.Hanif93)
    An alternative approach would be to notice that \dfrac{1+\sin x}{\cos x} \equiv \dfrac{1+\cos (\frac{\pi}{2} -x)}{\sin (\frac{\pi}{2} - x)} \equiv \dfrac{2\cos^2 (\frac{\pi}{4} - \frac{x}{2})}{2\sin (\frac{\pi}{4} - \frac{x}{2})\cos (\frac{\pi}{4} - \frac{x}{2})}, which after a bit of cancelling leads to logs directly.
    :cry2: thanks for that.
    I'd stick with my method though
 
 
 
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