How do you integrate $\int \frac{1}{\sqrt{x^{2} - 9}}$ $int 1/sqrt(x^2-9)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{x^{2} - 9}}$ $int 1/sqrt(x^2-9)$ by trigonometric substitution?

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Answer 1 · 2016-12-11T19:02:56

$\int \frac{d x}{\sqrt{x^{2} - 9}} = log ∣ ∣ ∣ \frac{x}{3} + \frac{1}{3} \sqrt{x^{9} - 9} ∣ ∣ ∣$ $int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|$

Explanation:

Substitute:

$x = 3 sec t$ $x=3sect$
$d x = 3 sec t tan t d t$ $dx = 3sect tant dt$

and you have for $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$ :

$\int \frac{d x}{\sqrt{x^{2} - 9}} = \int \frac{3 sec t tan t d t}{\sqrt{9 {sec}^{2} t - 9}} = \int \frac{sec t tan t d t}{\sqrt{\frac{1}{{cos}^{2} t} - 1}} = \int \frac{sec t tan t d t}{\sqrt{\frac{1 - {cos}^{2} t}{{cos}^{2} t}}} = \int \frac{sec t tan t d t}{\sqrt{\frac{{sin}^{2} t}{{cos}^{2} t}}} = \int \frac{sec t tan t d t}{\sqrt{{tan}^{2} t}}$ $int dx/sqrt(x^2-9) = int (3sect tant dt)/sqrt(9sec^2t-9)= int (sect tant dt)/sqrt(1/cos^2t-1) = int (sect tant dt)/sqrt((1-cos^2t)/cos^2t) = int (sect tant dt)/sqrt(sin^2t/cos^2t) =int (sect tant dt)/sqrt(tan^2t)$

In this interval for $tan t > 0$ $tant > 0$ , so:

$\int \frac{d x}{\sqrt{x^{2} - 9}} = \int \frac{sec t tan t d t}{tan t} = \int sec t d t = log | sec t + tan t |$ $int dx/sqrt(x^2-9) = int (sect tant dt)/tant =intsectdt= log |sect+ tant|$

To go back to x we note that:

$sec t = \frac{x}{3}$ $sect = x/3$

and $tan t = sin t \cdot sec t = sec t \sqrt{1 - \frac{1}{{sec}^{t} 2} t} = \frac{x}{3} \sqrt{1 - \frac{9}{x^{2}}} = \frac{1}{3} \sqrt{x^{9} - 9}$ $tant= sint*sect=sectsqrt(1-1/sec^t2t)= x/3sqrt(1-9/x^2)=1/3sqrt(x^9-9)$ .

Finally:

$\int \frac{d x}{\sqrt{x^{2} - 9}} = log ∣ ∣ ∣ \frac{x}{3} + \frac{1}{3} \sqrt{x^{9} - 9} ∣ ∣ ∣$ $int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|$

Answer 2 · 2016-12-11T19:19:39

Given: $\int \frac{1}{\sqrt{x^{2} - 9}} d x$ $int 1/sqrt(x^2 - 9)dx$

Let $x = 3 sec (u), then d x = 3 tan (u) sec (u) d u$ $x = 3sec(u)", then "dx = 3tan(u)sec(u)du$

Substitute this into the integral:

$\int \frac{3 tan (u) sec (u)}{\sqrt{{(3 sec (u))}^{2} - 9}} d u =$ $int (3tan(u)sec(u))/sqrt((3sec(u))^2 - 9)du =$

$\int \frac{3 tan (u) sec (u)}{\sqrt{9 {sec}^{2} (u) - 9}} d u =$ $int (3tan(u)sec(u))/sqrt(9sec^2(u) - 9)du =$

$\int \frac{3 tan (u) sec (u)}{3 \sqrt{{sec}^{2} (u) - 1}} d u =$ $int (3tan(u)sec(u))/(3sqrt(sec^2(u) - 1))du =$

$\int \frac{tan (u) sec (u)}{\sqrt{{sec}^{2} (u) - 1}} d u =$ $int (tan(u)sec(u))/(sqrt(sec^2(u) - 1))du =$

$\int \frac{tan (u) sec (u)}{\sqrt{{tan}^{2} (u)}} d u =$ $int (tan(u)sec(u))/(sqrt(tan^2(u)))du =$

$\int \frac{tan (u) sec (u)}{tan (u)} d u =$ $int (tan(u)sec(u))/tan(u)du =$

This integral is in any good list of integrals:

$\int sec (u) d u = ln (tan (u) + sec (u)) + C$ $int sec(u)du = ln(tan(u) + sec(u)) + C$

$\int \frac{1}{\sqrt{x^{2} - 9}} d x = ln (\sqrt{{sec}^{2} (u) - 1} + sec (u)) + C$ $int 1/sqrt(x^2 - 9)dx = ln(sqrt(sec^2(u) - 1) + sec(u)) + C$

Substitute $\frac{x}{3}$ $x/3$ for $sec (u)$ $sec(u)$ :

$\int \frac{1}{\sqrt{x^{2} - 9}} d x = ln (\sqrt{\frac{x^{2}}{9} - 1} + \frac{x}{3}) + C$ $int 1/sqrt(x^2 - 9)dx = ln(sqrt(x^2/9 - 1) + x/3) + C$

Answer 3 · 2016-12-11T21:00:47

${cosh}^{- 1} ∣ ∣ \frac{x}{3} ∣ ∣ + C$ $cosh^{-1)|x/3|+C$

Explanation:

Substitute $x = 3 cosh u$ $x=3cosh u$ , $d x = 3 sinh u d u$ $dx=3sinh u du$ and use ${cosh}^{2} u - {sinh}^{2} u = 1$ $cosh^2u -sinh^2u = 1$ to get the integral as being $u$ $u$ , which is ${cosh}^{- 1} (\frac{x}{3})$ $cosh^{-1}(x/3)$
which can be written as $ln ∣ ∣ x + \sqrt{x^{2} - 9} ∣ ∣ + C$ $ln|x+sqrt(x^2-9)|+C$ which is equivalent to
$ln ∣ ∣ ∣ (\frac{x}{3} + (\frac{1}{3}) \sqrt{x^{2} -} 9) ∣ ∣ ∣$ $ln |(x/3+(1/3)sqrt(x^2-)9)|$ given above, as the two answers differ by a constant of $ln 3$ $ln 3$ which is subsumed into the constant of integration.

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How do you integrate $\int \frac{1}{\sqrt{x^{2} - 9}}$ $int 1/sqrt(x^2-9)$ by trigonometric substitution?

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