# Math Help - trig integration

1. ## trig integration

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.

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

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

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

5. 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

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

6. 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.

7. 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.

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