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

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  • Thread starter karush
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In summary, the integral $\displaystyle \int\sqrt{{x}^{2}-1} \ dx$ can be evaluated using the substitution $\displaystyle x = \cosh{(t)}$, which simplifies the integral to $\displaystyle \int\sinh^2{(t)}\,dt$. This can be further simplified using hyperbolic identities to give the final result of $\displaystyle \frac{1}{2}\left[ x\,\sqrt{x^2 - 1} - \textrm{arcosh}\,(x) \right] + C$.
  • #1
karush
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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
 
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  • #2
karush said:
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...
 
  • #3
Why would this be any easier?
$$\displaystyle \int\sqrt{cosh^2 {(t) }-1 }\ \ sinh(t)\ dt $$
 
  • #4
\(\displaystyle \int\sqrt{\cosh^2t-1}\sinh t\,dt=\int\sinh^2t\,dt\)

Now use

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

Expand its square and integrate term by term. Of course, there may be other methods. :)
 
  • #5
$$
\displaystyle
\int\frac{e^{2t}}{4} \ dt
+
\int\frac{1}{4 e^{2t}} \ dt
-
\int\frac{1}{2} \ dt
=
\frac{e^{2t}}{8}
+
\frac{-e^{-2 t}}{8}
+
\frac{t}{2}
+
C
$$
I HOPE 😍 TI gave alternative answer??
$$
\displaystyle
\frac{x\sqrt{{x}^{2}-1}}{2}
-
\frac{\ln\left({\sqrt{{x}^{2}-1}+x}\right)}{2}
+
C
$$
 
Last edited:
  • #6
You would have to back-substitute for $t$ to get the anti-derivative as a function of $x$...:D
 
  • #7
So if
$$
\displaystyle
x=cosh(t)
$$
Then $t=?? $
TI says $t=-1$ 😫
 
  • #8
karush said:
$$
\displaystyle
\int\frac{e^{2t}}{4} \ dt
+
\int\frac{1}{4 e^{2t}} \ dt
-
\int\frac{1}{2} \ dt
=
\frac{e^{2t}}{8}
+
\frac{-e^{-2 t}}{8}
+
\frac{t}{2}
+
C
$$
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).
 
Last edited:
  • #9
Not sure why did you introduce $sinh(2t)$
 
  • #10
greg1313 said:
\(\displaystyle \int\sqrt{\cosh^2t-1}\sinh t\,dt=\int\sinh^2t\,dt\)

Now use

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

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...

$\displaystyle \begin{align*} \int{ \sinh^2{(t)}\,\mathrm{d}t} &= \frac{1}{2} \int{ \left[ \cosh{(2\,t)} - 1 \right] \,\mathrm{d}t } \\ &= \frac{1}{2}\,\left[ \frac{1}{2}\sinh{(2\,t)} - t \right] + C \\ &= \frac{1}{2}\,\left[ \cosh{(t)}\sinh{(t)} - t \right] +C \\ &= \frac{1}{2}\,\left[ \cosh{(t)}\,\sqrt{ \cosh^2{(t)} - 1 } - t \right] + C \\ &= \frac{1}{2}\,\left[ x\,\sqrt{x^2 - 1} - \textrm{arcosh}\,(x) \right] + C \end{align*}$
 
  • #11
ok. I see your point
I have never worked with hyperbolic identities

one thing I really like about MHB
 
  • #12
\(\displaystyle \dfrac14\dfrac{e^{2t}-e^{-2t}}{2}-\dfrac{t}{2}\)

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

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

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

-----

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

\(\displaystyle 2y=e^x+e^{-x}\)

\(\displaystyle 2ye^x=e^{2x}+1\)

\(\displaystyle e^{2x}-2ye^x+1=0\)

Quadratic formula:

\(\displaystyle e^x=\dfrac{2y+\sqrt{4y^2-4}}{2}=y+\sqrt{y^2-1}\)

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

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

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

-----

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

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

1. What is the meaning of "8.3.3 int sqrt (x^2-1) dx"?

The notation "8.3.3 int sqrt (x^2-1) dx" represents the definite integral of the square root of the quantity x squared minus one, with respect to the variable x, evaluated from 8 to 3.

2. How is the integral of the square root of a function evaluated?

To evaluate the integral of the square root of a function, you would first use the power rule to rewrite the function as a power of 1/2. Then, you would add one to the exponent and divide the result by the new exponent. Finally, you would evaluate the resulting function at the upper and lower limits and subtract the lower limit from the upper limit.

3. What is the purpose of the "dx" at the end of the integral notation?

The "dx" at the end of the integral notation represents the variable with respect to which the integration is being performed. In this case, the integral is being evaluated with respect to the variable x.

4. Can the definite integral be evaluated without knowing the function?

Yes, the definite integral can be evaluated without knowing the function by using numerical integration methods such as the trapezoidal rule or Simpson's rule. However, these methods may not always provide an exact solution and may require a large number of calculations.

5. What is the significance of the limits 8 and 3 in the integral notation?

The limits 8 and 3 represent the upper and lower limits of integration, respectively. These limits define the range over which the integration is being performed and determine the area under the curve being integrated.

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