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    How do you integrate tanx?

    Also, suppose we had
    \int (1/(4-5h))\ dx

    now the answer would simply be [latex]\ln 4-5h[latex]
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    How do you integrate tanx?



    Also, suppose we had
    \int \frac{1}{(4-5h)}\dx

    now the answer would simply be \frac{-1}{5}\ \ln 4-5h

    But why would it be wrong to multiply the numerator and denominator by -5 so that we have :
    \int \frac{-5}{-5(4-5h))}\ dx

    So we could factor out the -5 from the denominator and then we'd just have to integrate \frac{1}{(4-5h)} since this would be a case of the reverse chain rule.
    But this answer seems to have that extra factor of \frac{-1}{5}\
    Where do I go wrong in approaching the problem this way?
    \frac{1}{(4-5h)}
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    Write tanx= sinx/cosx
    Use the result: integral of f '(x) / f(x) = ln|f(x)| +c
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    to integrate tan(x), re-write it as sin(x)/cos(x), then use the substitution u=cos(x).

    With the integral, assuming that dx is meant to be dh (i.e. a typo), you're not quite there, since the top isn't quite the derivative of the bottom. What you need to do is re-arrange to get:
     \int \frac{1}{4-5h} dh = \frac{1}{-5} \int \frac{-5}{4-5h} dh=-\frac{1}{5} ln(4-5h)+c.
    If it isn't a typo, then essentially you just treat it as a constant, so you'd get \int \frac{1}{4-5h}\ dx = \frac{x}{4-5h} +c
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    -_- this is a double post :/
    and with regards to what vc94 said, don't forget the -
 
 
 
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Updated: April 4, 2011
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