MHB What is the integral of $\sqrt{x^2-1}$? What is the integral of $\sqrt{x^2-1}$?

[karush]

Evaluate
$$\displaystyle \int\sqrt{{x}^{2}-1} \ dx$$
First the indenitly of $\tan^2 \left({x}\right)=\sec^2 \left({x}\right)-1$ fits the expression in the radical

But not sure how to set up the substitution

 

[Prove It]

Evaluate
$$\displaystyle \int\sqrt{{x}^{2}-1} \ dx$$
First the indenitly of $\tan^2 \left({x}\right)=\sec^2 \left({x}\right)-1$ fits the expression in the radical

But not sure how to set up the substitution

If you were going to substitute $\displaystyle \begin{align*} x = \sec{(\theta)} \end{align*}$ then you need to have $\displaystyle \begin{align*} \mathrm{d}x = \sec{(\theta)}\tan{(\theta)}\,\mathrm{d}\theta \end{align*}$, making another difficult (though not impossible) integral.

The substitution $\displaystyle \begin{align*} x = \cosh{(t)} \implies \mathrm{d}x = \sinh{(t)}\,\mathrm{d}t \end{align*}$ will be easier...

 

[karush]

Why would this be any easier?
$$\displaystyle \int\sqrt{cosh^2 {(t) }-1 }\ \ sinh(t)\ dt $$

 

[Greg]

$$\int\sqrt{\cosh^2t-1}\sinh t\,dt=\int\sinh^2t\,dt$$

Now use

$$\sinh(t)=\dfrac{e^t-e^{-t}}{2}$$

Expand its square and integrate term by term. Of course, there may be other methods. :)

 

[karush]

$$
\displaystyle
\int\frac{e^{2t}}{4} \ dt
+
\int\frac{1}{4 e^{2t}} \ dt
-
\int\frac{1}{2} \ dt
=
\frac{e^{2t}}{8}
+
\frac{-e^{-2 t}}{8}
+
\frac{t}{2}
+
C
$$
I HOPE TI gave alternative answer??
$$
\displaystyle
\frac{x\sqrt{{x}^{2}-1}}{2}
-
\frac{\ln\left({\sqrt{{x}^{2}-1}+x}\right)}{2}
+
C
$$

 

[MarkFL]

You would have to back-substitute for $t$ to get the anti-derivative as a function of $x$...:D

 

[karush]

So if
$$
\displaystyle
x=cosh(t)
$$
Then $t=?? $
TI says $t=-1$

 

[Greg]

$$
\displaystyle
\int\frac{e^{2t}}{4} \ dt
+
\int\frac{1}{4 e^{2t}} \ dt
-
\int\frac{1}{2} \ dt
=
\frac{e^{2t}}{8}
+
\frac{-e^{-2 t}}{8}
+
\frac{t}{2}
+
C
$$
I HOPE TI gave alternative answer??
$$
\displaystyle
\frac{x\sqrt{{x}^{2}-1}}{2}
-
\frac{\ln\left({\sqrt{{x}^{2}-1}+x}\right)}{2}
+
C
$$

I'm assuming you've made a typo.

Since the original integral is indefinite we must back-substitute.

What you've got is

$$\dfrac14\dfrac{e^{2t}-e^{-2t}}{2}-\dfrac{t}{2}$$

If $x=\cosh(t)$ then $t=\cosh^{-1}(x)$

Now, given $\sinh(2t)=2\sinh(t)\cosh(t)$ and $\sinh(\cosh^{-1}(t))=\sqrt{t^2-1}$ can you complete the problem?

(We'll get to what your calculator says soon. Don't worry about that).

 

[karush]

Not sure why did you introduce $sinh(2t)$

 

[Prove It]

$$\int\sqrt{\cosh^2t-1}\sinh t\,dt=\int\sinh^2t\,dt$$

Now use

$$\sinh(t)=\dfrac{e^t-e^{-t}}{2}$$

Expand its square and integrate term by term. Of course, there may be other methods. :)

In my opinion, since the original substitution was given in terms of a hyperbolic function (i.e. not in its exponential form) it would be easier to work with the hyperbolic forms...

$\displaystyle \begin{align*} \int{ \sinh^2{(t)}\,\mathrm{d}t} &= \frac{1}{2} \int{ \left[ \cosh{(2\,t)} - 1 \right] \,\mathrm{d}t } \\ &= \frac{1}{2}\,\left[ \frac{1}{2}\sinh{(2\,t)} - t \right] + C \\ &= \frac{1}{2}\,\left[ \cosh{(t)}\sinh{(t)} - t \right] +C \\ &= \frac{1}{2}\,\left[ \cosh{(t)}\,\sqrt{ \cosh^2{(t)} - 1 } - t \right] + C \\ &= \frac{1}{2}\,\left[ x\,\sqrt{x^2 - 1} - \textrm{arcosh}\,(x) \right] + C \end{align*}$

 

[karush]

ok. I see your point
I have never worked with hyperbolic identities

one thing I really like about MHB

 

[Greg]

$$\dfrac14\dfrac{e^{2t}-e^{-2t}}{2}-\dfrac{t}{2}$$

$$\dfrac14\sinh(2t)-\dfrac{t}{2}=\dfrac12\left(\sinh(t)\cosh(t)-t\right)$$

$$\dfrac12\left(\sinh\left(\cosh^{-1}\right)\cosh\left(\cosh^{-1}(x)\right)-\cosh^{-1}(x)\right)$$

$$\dfrac12\left(x\sqrt{x^2-1}-\cosh^{-1}(x)\right)$$

-----

$$\cosh(x)=y=\dfrac{e^x+e^{-x}}{2}$$

$$2y=e^x+e^{-x}$$

$$2ye^x=e^{2x}+1$$

$$e^{2x}-2ye^x+1=0$$

Quadratic formula:

$$e^x=\dfrac{2y+\sqrt{4y^2-4}}{2}=y+\sqrt{y^2-1}$$

$$x=\ln\left(\sqrt{y^2-1}+y\right)$$

(Substituting cosh(x) for y in the line above gives reason for choosing the positive root).

$$\Rightarrow \cosh^{-1}x=\ln\left(\sqrt{x^2-1}+x\right)$$

-----

$$\int\sqrt{x^2-1}\,\mathrm dx=\dfrac12\left(x\sqrt{x^2-1}-\ln\left(\sqrt{x^2-1}+x\right)\right)+C$$