trig integration

**DivideBy0** · December 9th 2007, 09:46 PM

After I complete $\int \frac{\sqrt{x^2-9}}{x}\,dx$ by sec substitution I get

$3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C$ .

But this doesn't match the calculator's answer:

$3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C$

After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

Then the second question is $f(x) = 3$ . Find the equation.

**curvature** · December 9th 2007, 09:58 PM

Originally Posted by DivideBy0

After I complete $\int \frac{\sqrt{x^2-9}}{3}\,dx$ by sec substitution I get

$3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C$ .

But this doesn't match the calculator's answer:

$3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C$

After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

Then the second question is $f(x) = 3$ . Find the equation.

I guess the right integral is: $\int \frac{\sqrt{x^2-9}}{x}\,dx$

**DivideBy0** · December 9th 2007, 09:59 PM

Woops, sorry, you're right that's what I meant

**Soroban** · December 9th 2007, 10:16 PM

Hello, DivideBy0!

After I complete $\int \frac{\sqrt{x^2-9}}{x}\,dx$ by sec substitution I get: . $3\sec^{-1}\! \left(\frac{x}{3}\right)-\sqrt{x^2-3}+C$ .

But this doesn't match the calculator's answer: / $3\tan^{-1}\!\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C$

The answers are equivalent.

Let $\theta \:=\:\sec^{-1}\!\left(\frac{x}{3}\right)$
Then we have: . $\sec\theta \:=\:\frac{x}{3} \:=\:\frac{hyp}{adj}$

$\theta$ is in a right triangle with: $adj = 3,\;hyp = x$

. . Using Pythagorus: . $opp = \sqrt{x^2-9}$

So: . $\tan\theta \:=\:\frac{\sqrt{x^2-9}}{3}$
Hence:. . $\theta \:=\:\tan^{-1}\!\left(\frac{\sqrt{x^2-9}}{3}\right)$

See? . . . . . $\sec^{-1}\!\left(\frac{x}{3}\right) \;=\;\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)$

**curvature** · December 9th 2007, 10:19 PM

Originally Posted by DivideBy0

After I complete $\int \frac{\sqrt{x^2-9}}{x}\,dx$ by sec substitution I get

$3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C$ .

But this doesn't match the calculator's answer:

$3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C$

After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

Then the second question is $f(x) = 3$ . Find the equation.

the answar is
$-3\sec^{-1}\left(\frac{x}{3}\right)+\sqrt{x^2-9}+C$ .
or

$-3\cos^{-1}\left(\frac{3}{x}\right)+\sqrt{x^2-9}+C$ .

**DivideBy0** · December 9th 2007, 10:19 PM

Thanks Soroban for clearing it up a bit more... but when I differentiated on the calculator I still got

$\frac{d}{\,dx}\left(3\sec^{-1}(\frac{x}{3})-\sqrt{x^2-3}\right)=\frac{9|\frac{1}{x}|}{\sqrt{x^2-9}}-\frac{x}{\sqrt{x^2-3}}$

But the original expression obviously has no absolute value signs.

**Krizalid** · December 10th 2007, 05:28 AM

Originally Posted by DivideBy0

$\int \frac{\sqrt{x^2-9}}{x}\,dx$

You don't even need trig. sub. - this only requires a simple substitution.

Step #1: The make-up.

$\int {\frac{{\sqrt {x^2 - 9} }} {x}\,dx} = \int {\frac{{\sqrt {x^2 - 9} \cdot x}} {{x^2 }}\,dx} .$

Step #2: The substitution.

$u^2 = x^2 - 9 \implies u\,du = x\,dx,$

$\int {\frac{{\sqrt {x^2 - 9} }} {x}\,dx} = \int {\frac{{u^2 }} {{u^2 + 9}}\,du} .$

Step #3: Simple trick & mission almost-accomplished.

$\int {\frac{{u^2 }} {{u^2 + 9}}\,du} = \int {du} - 9\int {\frac{1} {{u^2 + 9}}\,du} = u - 3\arctan \frac{u} {3}+k.$

Step #4: Back substitute.

$\int {\frac{{\sqrt {x^2 - 9} }} {x}\,dx} = \sqrt {x^2 - 9} - 3\arctan \frac{{\sqrt {x^2 - 9} }} {3} + k.$

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