integral question..

**transgalactic** · February 28th 2009, 03:36 AM

$ \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\ $
$ z=\frac{1}{2}\tan x\\ $
$ dz=\frac{1}{2}\frac{dx}{\cos ^2 x} $
what to do next??

**mr fantastic** · February 28th 2009, 03:58 AM

Originally Posted by transgalactic

$ \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\ $
$ z=\frac{1}{2}\tan x\\ $
$ dz=\frac{1}{2}\frac{dx}{\cos ^2 x} $
what to do next??

Read this thread: http://www.mathhelpforum.com/math-he...functions.html

**transgalactic** · February 28th 2009, 04:34 AM

i am not using sec
only sin cos tan cot

its very complicated and with sec
so if i will put sec=1/cos then i ll get a very complicated expression
is there an easier way??

**Stonehambey** · February 28th 2009, 05:41 AM

is there an easier way??

You could try a hyperbolic substitution. In this example we're looking to reduce the sum of two squares to a single square, $z = \frac{1}{2}\sinh \theta$ will work here.

$z = \frac{1}{2}\sinh \theta \implies dz = \frac{1}{2}\cosh\theta\, d\theta$

So

$I = \int\frac{\frac{1}{2}\cosh\theta \, d\theta}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\f rac{1}{2}\right)^2\sinh\theta}} = \int d\theta = \theta + C$

We need to write this in terms of our original variable so

$z = \frac{1}{2}\sinh\theta \implies \theta = \sinh^{-1}(2z)$

So

$I = \sinh^{-1}(2z) + C$

This is also the answer maxima gives to the integral

**transgalactic** · February 28th 2009, 05:49 AM

i cant use hyperbolic function

**Stonehambey** · February 28th 2009, 06:11 AM

Why not?

Well if that's really the case then the substitution reduces your original integral to

$\int \sec x \, d x$

once you figure that out, draw a right angled triangle with $\tan x = 2z$ and use it to work out what any other trig functions in your answer are in terms of $z$

**mr fantastic** · February 28th 2009, 12:18 PM

Originally Posted by transgalactic

i am not using sec
only sin cos tan cot

its very complicated and with sec
so if i will put sec=1/cos then i ll get a very complicated expression
is there an easier way??

What do you mean you're not using sec?

What does the stuff in red say?:

Originally Posted by transgalactic

[snip]
$\int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}= {\color{red}\int \frac{dx}{\cos x}}$
[snip]
What to do next?

Unless my eyes need testing, I see sec. In which case, I have directed you to what to do next.

**mr fantastic** · February 28th 2009, 02:31 PM

Originally Posted by transgalactic

$ \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\ $
$ z=\frac{1}{2}\tan x\\ $
$ dz=\frac{1}{2}\frac{dx}{\cos ^2 x} $
what to do next??

This question may have been posted and answered at other websites that provide maths help.

**transgalactic** · March 3rd 2009, 09:07 AM

i didnt get an answer in a thread posted at another website

so help is needed

**Krizalid** · March 3rd 2009, 11:01 AM

http://www.mathhelpforum.com/math-he...tml#post274182

**mr fantastic** · March 3rd 2009, 03:33 PM

Originally Posted by transgalactic

i didnt get an answer in this thread
S.O.S. Mathematics CyberBoard :: View topic - integral question..

so help is needed

I gave you the answer in post #2. In post #3 you rejected what I posted. Now you're saying that you need the exact help I gave you that you had previously rejected.

Thread closed.

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