# Another integration question!

• Jul 26th 2006, 12:10 AM
margaritas
Another integration question!
Question: Using the substitution x = 1/y, or otherwise, find: (see attached)

Sorry for the picture quality, I took it off my phone.

It's x and in the bracket, x^2 you see there.

• Jul 26th 2006, 12:29 AM
earboth
Quote:

Originally Posted by margaritas
Question: Using the substitution x = 1/y, or otherwise, find: (see attached)

Sorry for the picture quality, I took it off my phone.
It's x and in the bracket, x^2 you see there.

Hello, margaritas,

do you mean: $\int{\frac{1}{x \cdot \sqrt{x^2-1}}dx}$

If so the result is $\arctan \left( \sqrt{x^2-1} \right)$

Steps to reach this result will follow.

Greetings

EB
• Jul 26th 2006, 12:54 AM
margaritas
Thanks for the help, EB, yep that's the question. But the answer as given by my tutor is supposedly: C - sin^(-1) (1/x).
• Jul 26th 2006, 10:27 AM
earboth
Quote:

Originally Posted by margaritas
Question: Using the substitution x = 1/y, or otherwise, find: (see attached)...

Hello, margaritas,
it's me again.

As you suggested, use $x=\frac{1}{y}\ \Longrightarrow\ \frac{dx}{dy}=-\frac{1}{y^2}$. So ${dx}=-\frac{1}{y^2}\cdot {dy}$.

Substitute the x and the dx and you'll get:

$\int{\frac{1}{\frac{1}{y}\cdot \sqrt{\frac{1}{y^2}-1}}\cdot \left(-\frac{1}{y^2}\right)\cdot dy}$. Rearrange:

$\int{\frac{y\cdot y}{\sqrt{1-y^2}}\cdot \left(-\frac{1}{y^2}\right) \cdot dy}$ = $-\int{\frac{1}{\sqrt{1-y^2}} \cdot dy}$

As you may see, this is $-\int{\frac{1}{\sqrt{1-y^2}} \cdot dy} = -\arcsin(y)+C=C-\arcsin\left(\frac{1}{x}\right)$

So the answer given in your book is OK - but my result is true too. There are integrals with different solutions. After a few steps of transformation, you'll reach my result. Have a look here:

Greetings

EB
• Jul 27th 2006, 06:24 AM
Soroban
Great work, EB!

It seems to invite Trig Substitution,
. . so I tried it . . . and found yet another solution.

$\int\frac{dx}{x\sqrt{x^2-1}}$

Let $x = \sec\theta\quad\Rightarrow\quad dx = \sec\theta\tan\theta\,d\theta\quad\Rightarrow\quad \sqrt{x^2 - 1} = \tan\theta$

Substitute: . $\int\frac{\sec\theta\tan\theta\,d\theta}{\sec \theta\tan\theta}\;=\;\int d\theta \;=\;\theta + C$

Back-substitute: . $\boxed{\text{arcsec }x + C}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I had: $\sec\theta = \frac{x}{1}$

$\theta$ is an angle is a right triangle with: $hyp = x,\;adj = 1$
Code:

                  *           x    *  |  ______             *    | √x² - 1         * θ      |       * - - - - - *             1

Your answer . $\arctan\sqrt{x^2-1}$ .is the same angle.

The other acute angle is: $\frac{\pi}{2} - \theta$ . and $\sin\left(\frac{\pi}{2} - \theta\right) = \frac{1}{x}$

Hence: . $\frac{\pi}{2} - \theta\:=\:\arcsin\left(\frac{1}{x}\right)\quad \Rightarrow\quad\theta \:= \:\frac{\pi}{2} - \arcsin\left(\frac{1}{x}\right)$

. . which verifies margaritas' solution.

• Jul 29th 2006, 04:52 AM
margaritas
Thanks for the help, guys! :)