1. ## integral question..

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

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

3. 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??

4. 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, $\displaystyle z = \frac{1}{2}\sinh \theta$ will work here.

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

So

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

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

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

So

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

This is also the answer maxima gives to the integral

5. i cant use hyperbolic function

6. Why not?

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

$\displaystyle \int \sec x \, d x$

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

7. 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]
$\displaystyle \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}= {\color{red}\int \frac{dx}{\cos x}}$
[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.

8. Originally Posted by transgalactic
$\displaystyle \int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\$
$\displaystyle z=\frac{1}{2}\tan x\\$
$\displaystyle dz=\frac{1}{2}\frac{dx}{\cos ^2 x}$
what to do next??
This question may have been posted and answered at other websites that provide maths help.

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

so help is needed

10. Originally Posted by transgalactic