Need help with an integral

**angrynapkin** · November 13th 2008, 11:49 PM

1 / sqrt(2x)

I tried splitting things up but I'm just running myself in circles at this point. Two methods I tried are:

pulling 1/sqrt(2) out so I get

1/sqrt(2) * 1/sqrt(x) <---But I can't integrate that

I also tried to rationalize by multiplying top and bottom by sqrt(2x) but I just end up with something I still can't integrate.

A really convoluted thing I did was change the sqrt(2x) to 2x^1/2

Separate 1/2 out of that so I get 1/2 * x^-1/2 and integrate from there. No luck, as I just end up with 2sqrt(x) / 2, which is not the right answer. Can anyone shed some light on this?

p.s. - sorry, I'm not quite sure how to use pretty print, u-sub is in the next section so i'm trying to do it the hard way first.

**ADARSH** · November 14th 2008, 12:14 AM

Hello,

Here's a clue
put $\sqrt{x} = t$
and go on

feel free to ask if there's is still trouble

Adarsh

**angrynapkin** · November 14th 2008, 12:29 AM

Adarsh,

Thank you for your reply, but I'm not quite sure I'm catching on.

I thought 1 / sqrt(2) * 1 / sqrt(x) was the the way to go, but no so sure.

**ADARSH** · November 14th 2008, 12:37 AM

put $\sqrt{x} = t$
differntiate both sides
$\frac{dx}{2\sqrt{x}}=dt$
$dx=2\sqrt{x}dt$
put this in
$\int{\frac{dx}{\sqrt{2x}}}$
you get
$\int{\sqrt{2}dt}$
so answer is
$=\sqrt{2}t + c$
$=2\sqrt{x}+c$

Adarsh

Thread: Need help with an integral

LinkBack

Thread Tools

Display

Need help with an integral

Thanks

..

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Evaluating a double integral to find its signle integral.

Search Tags