∫ (sec x)(sec x)(tan x)

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  1. sabre2th1's Avatar
    1 15-06-2012  14:41
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    ∫ (sec x)(sec x)(tan x)
    ∫ (sec x)(tan x) = sec x
    so isn't ∫ (sec x)(sec x)(tan x) = ∫(sec x)(tan x) x ∫(sec x) ?
  2. SecondHand's Avatar
    2 15-06-2012  14:45
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    Re: ∫ (sec x)(sec x)(tan x)
    No.

    \displaystyle \int sec^{2}x tan{x} = \frac{1}{2} tan^{2} x
  3. notnek's Avatar
    3 15-06-2012  14:47
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by sabre2th1)
    ∫ (sec x)(tan x) = sec x
    so isn't ∫ (sec x)(sec x)(tan x) = ∫(sec x)(tan x) x ∫(sec x) ?
    You can't do that with integrals. In general:

    \displaystyle \int f(x) g(x) \ dx \neq \int f(x) \ dx \times \int g(x) \ dx
  4. sabre2th1's Avatar
    4 15-06-2012  23:05
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by SecondHand)
    No.

    \displaystyle \int sec^{2}x tan{x} = \frac{1}{2} tan^{2} x

    (Original post by notnek)
    You can't do that with integrals. In general:

    \displaystyle \int f(x) g(x) \ dx \neq \int f(x) \ dx \times \int g(x) \ dx
    Ah.. I see. I am still confused on how:

    ∫[(2sec x)(sec x)(tan x)] = sec2x


    Thanks
  5. cpdavis's Avatar
    5 15-06-2012  23:08
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by sabre2th1)
    Ah.. I see. I am still confused on how:

    ∫[(2sec x)(sec x)(tan x)] = sec2x


    Thanks
    What's the differential of tanx? Now look at the differential of tan^2 x
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  6. raheem94's Avatar
    6 15-06-2012  23:11
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by sabre2th1)
    Ah.. I see. I am still confused on how:

    ∫[(2sec x)(sec x)(tan x)] = sec2x


    Thanks
    Its an integral of the form \displaystyle \int f'(x) [f(x)]^n \ dx

    For such integrals we try differentiating, [f(x)]^{n+1}

    Spoiler:
    Show

    \displaystyle u = sec^2 x \\ \frac{du}{dx} = 2 (secx) \times secx tanx = 2sec^2 x tanx
  7. ian.slater's Avatar
    7 16-06-2012  14:49
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by sabre2th1)
    Ah.. I see. I am still confused on how:

    ∫[(2sec x)(sec x)(tan x)] = sec2x


    Thanks
    You've been given two different right answers to your question. Differentiate either sec^(x) or tan^2(x) using the chain rule and you get the integrand you started with. You might reasonably ask 'which is it, then?'

    When you integrate you introduce an arbitrary constant. What's the difference between sec^2(x) and tan^2(x)?
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  8. sabre2th1's Avatar
    8 17-06-2012  18:33
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by cpdavis)
    What's the differential of tanx? Now look at the differential of tan^2 x

    (Original post by raheem94)
    Its an integral of the form \displaystyle \int f'(x) [f(x)]^n \ dx

    For such integrals we try differentiating, [f(x)]^{n+1}

    Spoiler:
    Show

    \displaystyle u = sec^2 x \\ \frac{du}{dx} = 2 (secx) \times secx tanx = 2sec^2 x tanx

    (Original post by ian.slater)
    You've been given two different right answers to your question. Differentiate either sec^(x) or tan^2(x) using the chain rule and you get the integrand you started with. You might reasonably ask 'which is it, then?'

    When you integrate you introduce an arbitrary constant. What's the difference between sec^2(x) and tan^2(x)?
    I've got it thanks alot !
  9. sabre2th1's Avatar
    9 17-06-2012  19:19
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by raheem94)
    Its an integral of the form \displaystyle \int f'(x) [f(x)]^n \ dx

    For such integrals we try differentiating, [f(x)]^{n+1}

    Spoiler:
    Show

    \displaystyle u = sec^2 x \\ \frac{du}{dx} = 2 (secx) \times secx tanx = 2sec^2 x tanx
    If you have ∫(sec2x)(tan2x), then how do you determine which [of (sec2x) and (tan2x)] is the f(x) and which is the f'(x)?
  10. ian.slater's Avatar
    10 17-06-2012  19:56
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by sabre2th1)
    If you have ∫(sec2x)(tan2x), then how do you determine which [of (sec2x) and (tan2x)] is the f(x) and which is the f'(x)?
    I've tried it both ways, and only one works.
  11. ztibor's Avatar
    11 17-06-2012  22:38
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    Re: ∫ (sec x)(sec x)(tan x)
    (Original post by ian.slater)
    I've tried it both ways, and only one works.
    \int 2 sec^2x\cdot tanx dx=tan^2x+C=tan^2x+1+C'=sec^2x+C '

    Given that
    sin^2x+cos^2x=1
    Divide both side by cos^2x (\neq 0)
    tan^2x+1=\frac{1}{cos^2x}
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