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anonymous

  • 5 years ago

complete the following integral or partial substitution with x tanx dx

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  1. dumbcow
    • 5 years ago
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    use integration by parts u = x dv = tanx du =dx v = -ln|cosx|

  2. dumbcow
    • 5 years ago
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    wait nevermind that wont work i dont know if this has a simple anti-derivative

  3. anonymous
    • 5 years ago
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    Do integration by parts, Take the derivative of x and integral of tanx =\[x \log|secx| - \int\limits_{}{}tanx\] =log|secs|(x-1)

  4. anonymous
    • 5 years ago
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    That should be secx*

  5. anonymous
    • 5 years ago
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    Sorry did a mistake, its not that simple, will think on it!

  6. anonymous
    • 5 years ago
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    = uv - integral v du so now all thats left is finding integral of ln(secx)

  7. anonymous
    • 5 years ago
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    let u= sec(x) , du = sec(x)tan(x) dx --> dx = du/ sec(x)tan(x)

  8. anonymous
    • 5 years ago
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    so integral ln(u) du / sec(x) tan(x) = \[\frac{ \ln(u) du }{u \sqrt{u^2-1}} \]

  9. anonymous
    • 5 years ago
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    yes , hmm, I dont think you can go anywhere there

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