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