Integration Query! (1 Viewer)

[lookoutastroboy]

Hey guys,

In the 2005 HSC paper, for Q1d. - where does the 1/2 come from? For any sort of 1/sqrt (x^2 + a^2)) or 1/sqrt (x^2 - a^2) - i always get it into the standard form - but book has different answers like this one... Could someone please explain this?






Cheers,
L.A.

 

[K4M1N3]

First factorise out a 4 from UNDER the square root. Since 4 is a square number it can be taken out of the square root as a 2. Since we take the 2 out from the denominator, its actually a 1/2...which can be taken out of the whole integrand.

Then you would have something like x^2 - 1/4 under the square root, which you can apply to the standard integral.

i.e.

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x@plus;\sqrt{x^2 - \frac{1}{4}} |" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x+\sqrt{x^2 - \frac{1}{4}} |" title="\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x+\sqrt{x^2 - \frac{1}{4}} |" /></a>

 

[lookoutastroboy]

First factorise out a 4 from UNDER the square root. Since 4 is a square number it can be taken out of the square root as a 2. Since we take the 2 out from the denominator, its actually a 1/2...which can be taken out of the whole integrand.

Then you would have something like x^2 - 1/4 under the square root, which you can apply to the standard integral.

i.e.

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x@plus;\sqrt{x^2 - \frac{1}{4}} |" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x+\sqrt{x^2 - \frac{1}{4}} |" title="\int \frac{dx}{\sqrt{4x^2 - 1}} \\ \int \frac{dx}{\sqrt{4(x^2 - \frac{1}{4})}}\\ \int \frac{dx}{2\sqrt{(x^2 - \frac{1}{4})}}\\ \frac{1}{2}\int \frac{dx}{\sqrt{(x^2 - \frac{1}{4})}}\\ =\frac{1}{2}ln|x+\sqrt{x^2 - \frac{1}{4}} |" /></a>

That's correct and what I'm always doing. But how come the answer has 2x in the ln part? Thats what I dont understand...

 

[kooliskool]

If you use substitution of u=2x, then you will get the book's answer

 

[lookoutastroboy]

If you use substitution of u=2x, then you will get the book's answer

omg thanks a lot! i get it now, finally...

 

[funnytomato]

also both answers are correct, with constants

since ln |2x+sqrt(4x^2-1)| + c = ln 2*|x+sqrt(x^2-1/4)| + c =ln |x+sqrt(x^2-1/4)| + ln2 + c =ln |x+sqrt(x^2-1/4)|+ k

where c and k are arbitary constants , k = c+ln2