Question

find integration of (secx-1)^1/2

Answer 1

do you mean\[\int\limits\limits[ (\sec x)-1 ]^{(1/2)} dx ?\] or \[\int\limits\limits[ \sec( x-1) ]^{(1/2)} dx ?\]

Answer 2

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

Answer 3

\(if\) it's the first one

Answer 4

it must be\[\int\sqrt{\sec x-1}dx\]the other one is not really doable

Answer 5

damn! I scribbled this integral with the solution on the margin of my notebook and now I can't find it!!!

Answer 6

ok Fermat ;)

Answer 7

its first one

Answer 8

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

Answer 9

but whats the process i din't find any process in wolf

Answer 10

click "show steps"

Answer 11

can anyone check it for me?

Answer 12

i am not gettin step after taking (secx)-1=u

Answer 13

hey callisto your attachment is not opening properly whould you plz write or draw it bro

Answer 14

i prefer sis to bro lol

Answer 15

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

Answer 16

the 6th step is...?

Answer 17

you are correct in\[\int\sec xdx=\ln(|\sec x+\tan x|+C\]

Answer 18

@TuringTest
and what's wrong with the other part? I need to learn from mistakes

Answer 19

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?

Answer 20

i mean to do something like ∫(ax+c)^n dx and that's sth i don't know how to deal with

Answer 21

so i tried the derivative way..

Answer 22

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\[\int(\sec x-1)^{1/2}dx\neq\frac32(\sec x-1)^{3/2}+C\]?

Answer 23

and what should i do?

Answer 24

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

Answer 25

okay, thanks