i was asked to evaluate:

$\displaystyle

\int \sqrt{x^2 - 4} dx

$

using a substitution method involving hyperbolic or trig function.

ive seen some people use x = 2 sec(theta) .ive tried to integrate it but got stuck mid way.

pls help

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- Apr 13th 2008, 02:49 AMtasuketeintegration with substitution of trig/hyperbolic function
i was asked to evaluate:

$\displaystyle

\int \sqrt{x^2 - 4} dx

$

using a substitution method involving hyperbolic or trig function.

ive seen some people use x = 2 sec(theta) .ive tried to integrate it but got stuck mid way.

pls help - Apr 13th 2008, 03:04 AMMoo
Hello,

$\displaystyle cosh^2(x)-sinh^2(x)=1$

$\displaystyle (2cosh(x))^2-(2sinh(x))^2=4$

$\displaystyle (2cosh(x))^2-4=(2sinh(x))^2$

$\displaystyle \sqrt{(2cosh(x))^2-4}=2sinh(x)$

So $\displaystyle u=2cosh(x)$ seems to be a pretty good idea, plus, its antiderivative is known (sinh(x)) - Apr 13th 2008, 06:28 AMmr fantastic
$\displaystyle x = 2 \sec t \Rightarrow \frac{dx}{dt} = -2 \frac{\sin t}{\cos^2 t} \Rightarrow dx = -2 \tan t \, \sec t \, dt$.

Then the integral becomes

$\displaystyle -4 \int \tan t (\tan t \, \sec t) \, dt = -4 \int \tan^2 t \, \sec t \, dt$

$\displaystyle = \int (\sec^2 t - 1) \, \sec t \, dt = \int \sec^3 t \, dt - \int \sec t \, dt$.

The second integral can be found in several ways. The result is $\displaystyle \ln |\sec t + \tan t|$.

The first requires integration by parts and then the above result.

Then you have to substitute back that $\displaystyle x = 2 \sec t$.

The hyperbolic substitution is easier. - Apr 13th 2008, 09:48 AMtasuketesomething went wrong
- Apr 13th 2008, 09:57 AMtasukete
$\displaystyle x = 2cosh(t)

$

$\displaystyle

dx/d(t) = 2sinh(t)

$

$\displaystyle

dx = 2sinh(t) d(t)

$

$\displaystyle

\int \sqrt{x^2 - 4} dx

$

$\displaystyle

= \int 2sinh(t)(2sinh(t)) d(t)

$

$\displaystyle

= 2\int 2sinh^2(t) d(t)

$

$\displaystyle

cosh(2t) = 1 + 2sinh^2(t)

$

$\displaystyle

2sinh^2(t) = cosh(2t) - 1

$

$\displaystyle

= 2\int 2sinh^2(t) = 2 \int cosh (2t) - 1 d(t)

$

$\displaystyle

= 2 [sinh(t) - t]

$

something went wrong..:S - Apr 13th 2008, 10:13 AMMoo
The derivative for x is sinh(t), not -sinh(t) ;)

- Apr 13th 2008, 10:16 AMtasukete