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    Hi,

    I have to integrate tan, and I've used the two different approaches but neither give me the right answer. Below is my working, please could someone tell me what I've done wrong on either/both approaches?


    \displaystyle \int_0^{0.4} \frac {2tanx}{1-tan^2x} . So when I saw this I thought it was quite a nice one as it related directly to the identity tan2x with no rearranging needed.


    \displaystyle \int_0^{0.4} tan2x

    So I used the  \frac{1}{2} ln(sec2x) approach first but that give me a really small value that was way off:


     (\frac{1}{2} ln[sec0.8]) -  (\frac{1}{2} ln[sec0]) = 0.000049 which is way off.

    So then I converted tan into sin/cos and integrated that way:

     \displaystyle \int_0^{0.4} \frac{sin2x}{cos2x} = \frac{-\frac{1}{2}cos2x}{\frac{1}{2}sin  2x} = -71.62

    The answer in the book is 0.181
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    (Original post by marcsaccount)
    Hi,

    I have to integrate tan, and I've used the two different approaches but neither give me the right answer. Below is my working, please could someone tell me what I've done wrong on either/both approaches?


    \displaystyle \int_0^{0.4} \frac {2tanx}{1-tan^2x} . So when I saw this I thought it was quite a nice one as it related directly to the identity tan2x with no rearranging needed.


    \displaystyle \int_0^{0.4} tan2x

    So I used the  \frac{1}{2} ln(sec2x) approach first but that give me a really small value that was way off:


     (\frac{1}{2} ln[sec0.8]) -  (\frac{1}{2} ln[sec0]) = 0.000049 which is way off.

    So then I converted tan into sin/cos and integrated that way:

     \displaystyle \int_0^{0.4} \frac{sin2x}{cos2x} = \frac{-\frac{1}{2}cos2x}{\frac{1}{2}sin  2x} = -71.62

    The answer in the book is 0.181
    Get yourself into radians (rather than degrees) and you'll be fine :L
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    (Original post by alexmufc1995)
    Get yourself into radians (rather than degrees) and you'll be fine :L
    Oh dear! The time and pain! Thanks for pointing it out!

    Marc
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    (Original post by alexmufc1995)
    Get yourself into radians (rather than degrees) and you'll be fine :L
     \int \frac{\sin 2x}{\cos 2x} dx is NOT what you have put. You do not integrate the numerator and denominator separately when integrating a rational function. This would actually be a logarithmic integral.
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    interestingly though, only the approach using natural logs gives the correct answer. Why wouldn't the sin/cos method work? Is my working above correct?
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    (Original post by brianeverit)
     \int \frac{\sin 2x}{\cos 2x} dx is NOT what you have put. You do not integrate the numerator and denominator separately when integrating a rational function. This would actually be a logarithmic integral.

    thanks
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    (Original post by marcsaccount)
    interestingly though, only the approach using natural logs gives the correct answer. Why wouldn't the sin/cos method work? Is my working above correct?
    Because the numerator is the differential of the denominator (with a minus sign), so the integral is a natural log.

    You can't just integrate the denominator and numerator, you would have to do it by parts. Although I doubt you would get an answer, at least not quickly.
 
 
 
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