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    Q: ∫1+sinx/cosx dx ; u = sin x

    I can't get it into something that i can integrate..

    I get upto ∫1+u/1-u² but i'm stuck on what to do next, can someone help me please?
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    Try splitting it into int(1/(1-u^2) + int(u/(1-u^2))

    Sorry for the untidyness of the maths above, I'll read the FAQ on doing maths symbols.
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    Tom, are you using the Quick reply box? If so, click 'Reply' (in the post bit) or 'Go advanced' in the quick reply box, and you'll see a full set of Maths symbols. =)

    Btw, I would've thought it's easier just to say (1-u2) = (1+u)(1-u), and cancel the (1+u) term.
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    (Original post by mockel)
    Tom, are you using the Quick reply box? If so, click 'Reply' (in the post bit) or 'Go advanced' in the quick reply box, and you'll see a full set of Maths symbols. =)

    Btw, I would've thought it's easier just to say (1-u2) = (1+u)(1-u), and cancel the (1+u) term.
    Yeah that probably would be easier, oh well :P

    I thought: arctan u + ln(1-u²)/2 + c was a little complicated, it's probably wrong too.

    Thanks for the tip on doing maths symbols.
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    (Original post by mockel)
    Tom, are you using the Quick reply box? If so, click 'Reply' (in the post bit) or 'Go advanced' in the quick reply box, and you'll see a full set of Maths symbols. =)

    Btw, I would've thought it's easier just to say (1-u2) = (1+u)(1-u), and cancel the (1+u) term.
    Oh yeh, that works.. Thanks.... again!
    I need to train my brain to work like yours :rolleyes:
 
 
 
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