# integration with substitution of trig/hyperbolic function

• April 13th 2008, 02:49 AM
tasukete
integration with substitution of trig/hyperbolic function

$
\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
• April 13th 2008, 03:04 AM
Moo
Hello,

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

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

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

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

So $u=2cosh(x)$ seems to be a pretty good idea, plus, its antiderivative is known (sinh(x))
• April 13th 2008, 06:28 AM
mr fantastic
Quote:

Originally Posted by tasukete

$
\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

$x = 2 \sec t \Rightarrow \frac{dx}{dt} = -2 \frac{\sin t}{\cos^2 t} \Rightarrow dx = -2 \tan t \, \sec t \, dt$.

Then the integral becomes

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

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

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

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

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

The hyperbolic substitution is easier.
• April 13th 2008, 09:48 AM
tasukete
something went wrong
Quote:

Originally Posted by Moo
Hello,

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

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

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

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

So $u=2cosh(x)$ seems to be a pretty good idea, plus, its antiderivative is known (sinh(x))

i dont know where i went wrong... somehow ended up with -ve intergral.. my final answer came to be -2sinh(theta) + 2theta (Headbang)
• April 13th 2008, 09:57 AM
tasukete
$x = 2cosh(t)
$

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

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

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

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

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

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

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

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

$
= 2 [sinh(t) - t]
$

something went wrong..:S
• April 13th 2008, 10:13 AM
Moo
The derivative for x is sinh(t), not -sinh(t) ;)
• April 13th 2008, 10:16 AM
tasukete
Quote:

Originally Posted by Moo
The derivative for x is sinh(t), not -sinh(t) ;)

oh my..i need sleep (Crying) thnx heaps ^^