Question

∫{dy/y(4y^2-)^(1/2)

Find the definite integral

Answer 1

-1 is under the square root too?

Answer 2

\[\int\limits_{}^{}{\frac{dy}{y \sqrt{4y^{2}-1}}}\]

Answer 3

thats the correct way

Answer 4

I am not sure, thinking

Answer 5

yeah this is a tricky one...

Answer 6

I dont know, sorry :(

Answer 7

I am stuck

Answer 8

I can help you :D

Answer 9

Here is the help! :)

Answer 10

:D Let me write it down and then I'll type it up xP

Answer 11

ok thank you so much! andras...alright...i'll wait for your response malevolence

Answer 12

This is the way I did it. I noticed that you have a sqrt in the form of sqrt(u^2-a^2). So rewrite it this way:
\[\frac{1}{2}\int\limits \frac{dy}{y \sqrt{y^2-\frac{1}{4}}}\]
Use the substitution sec(x)=y
so: sec(x)tan(x)dx=dy
and tan(x)=sqrt(y-(1/4))
Rewriting then you have:
\[\frac{1}{2} \int\limits \frac{\sec(x)\tan(x)dx}{\sec(x)\tan(x)}=\frac{1}{2}\int\limits dx=\frac{x}{2}\]
But x=arcsec(y)
So the integral is:
\[\frac{\sec^{-1}(y)}{2}+C\]
However, wolfram did it slightly differently but I'm assuming they are equivalent.
http://www.wolframalpha.com/input/?i=integral+of+dx%2F(x(sqrt(4x^2-1)))

Answer 13

wow..okay...so if we never learn that 1/(u^2+1) is tan inverse?

thats the only way to do it?

and thanks so much malevolence...

Answer 14

The trig substitution that I did above is who I know how to do it. The u-sub that wolfram did is crafty. I feel as though they won't be any different however. Try your definite integral and see I suppose.

Answer 15

And you're welcome :)