# integration with substitution of trig/hyperbolic function

Printable View

• Apr 13th 2008, 02:49 AM
tasukete
integration 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 AM
Moo
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 AM
mr fantastic
Quote:

Originally Posted by tasukete
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

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

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 AM
tasukete
something went wrong
Quote:

Originally Posted by Moo
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))

i dont know where i went wrong... somehow ended up with -ve intergral.. my final answer came to be -2sinh(theta) + 2theta (Headbang)
• Apr 13th 2008, 09:57 AM
tasukete
$\displaystyle x = 2cosh(t)$
$\displaystyle dx/d(t) = 2sinh(t)$
$\displaystyle dx = 2sinh(t) d(t)$
$\displaystyle \int \sqrt{x^2 - 4} dx$
$\displaystyle = \int 2sinh(t)(2sinh(t)) d(t)$
$\displaystyle = 2\int 2sinh^2(t) d(t)$

$\displaystyle cosh(2t) = 1 + 2sinh^2(t)$
$\displaystyle 2sinh^2(t) = cosh(2t) - 1$
$\displaystyle = 2\int 2sinh^2(t) = 2 \int cosh (2t) - 1 d(t)$
$\displaystyle = 2 [sinh(t) - t]$

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