anonymous
  • anonymous
find integration of (secx-1)^1/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Callisto
  • Callisto
do you mean\[\int\limits\limits[ (\sec x)-1 ]^{(1/2)} dx ?\] or \[\int\limits\limits[ \sec( x-1) ]^{(1/2)} dx ?\]
TuringTest
  • TuringTest
I see no better way than wolfram's http://www.wolframalpha.com/input/?i=+%28secx-1%29%5E1%2F2dx someone please show me a more elegant solution
TuringTest
  • TuringTest
\(if\) it's the first one

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TuringTest
  • TuringTest
it must be\[\int\sqrt{\sec x-1}dx\]the other one is not really doable
anonymous
  • anonymous
damn! I scribbled this integral with the solution on the margin of my notebook and now I can't find it!!!
TuringTest
  • TuringTest
ok Fermat ;)
anonymous
  • anonymous
its first one
TuringTest
  • TuringTest
what kills me is how the solution from wolf is\[2\tanh^{-1}(\sqrt{\sec x+1})+C\]and then says "which for restricted values is equivalent to (a bunch of ugly crap)" still, I see no better way than wolf's here, and it's not that hard
anonymous
  • anonymous
but whats the process i din't find any process in wolf
TuringTest
  • TuringTest
click "show steps"
Callisto
  • Callisto
can anyone check it for me?
1 Attachment
anonymous
  • anonymous
i am not gettin step after taking (secx)-1=u
anonymous
  • anonymous
hey callisto your attachment is not opening properly whould you plz write or draw it bro
Callisto
  • Callisto
i prefer sis to bro lol
1 Attachment
anonymous
  • anonymous
sorry callisto but both are wrong as in 1 we can't one variable out of integration and one in check at 6th step of your attachmenrt and the second one i didn't get it
Callisto
  • Callisto
the 6th step is...?
TuringTest
  • TuringTest
you are correct in\[\int\sec xdx=\ln(|\sec x+\tan x|+C\]
Callisto
  • Callisto
@TuringTest and what's wrong with the other part? I need to learn from mistakes
TuringTest
  • TuringTest
I'm not really sure what theorem you think youare applying in this step:\[\int(\sec x-1)^{1/2}dx={\frac32(\sec x-1)\over \int\sec xdx}+C\]that move really makes no sense to -why do you think you can divide by integral sec? -what happened to the 1/2 exponent on top? -where did 3/2 come from?
Callisto
  • Callisto
i mean to do something like ∫(ax+c)^n dx and that's sth i don't know how to deal with
1 Attachment
Callisto
  • Callisto
so i tried the derivative way..
TuringTest
  • TuringTest
I don't know what you mean the 'derivative way', but if you mean differentiation under the integral sign I think that technique is too advanced for you right now, and you have done it wrong. it looks like you tried to do this\[\int(\sec x-1)^{1/2}dx={\int(\sec x-1)^{1/2}\over \int\sec xdx}={\frac23(\sec x-1)^{3/2}\over \int\sec xdx}+C\]but the top cannot be integrated like that you also have no basis to divide the bottom by integral sec if you could just apply the power rule like that this problem would be a piece of cake, but you can't... do you see that\[\int(\sec x-1)^{1/2}dx\neq\frac32(\sec x-1)^{3/2}+C\]?
Callisto
  • Callisto
and what should i do?
TuringTest
  • TuringTest
as I say in the above posts, this seems to be a pretty difficult integral The only way I can see to do it is wolframs way, though I am sure there is a better one out there somewhere
Callisto
  • Callisto
okay, thanks

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