# Thread: Hyperbolic Integration by substitution

1. ## Hyperbolic Integration by substitution

Using the substitution $\displaystyle x = \frac{3}{sinh \theta}$, find the exact value of:

$\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{x\sqrt{x^2 + 9}} dx$

If $\displaystyle x = \frac{3}{sinh \theta}$, then $\displaystyle \frac{dx}{d\theta} = \frac{-3cosh \theta}{sinh^2 \theta}$.

Putting in the substitution, I get $\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{\frac{3}{sinh \theta}\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh^2 \theta} d\theta$.

From this I moved the $\displaystyle sinh \theta$ to the top of the first fraction and cancelled with one of the $\displaystyle sinh^2 \theta$ on the denominator of the second fraction to give me this.

$\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{3\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

I have no idea what to do next, would really appreciate some help?

2. Originally Posted by craig

Using the substitution $\displaystyle x = \frac{3}{sinh \theta}$, find the exact value of:

$\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{x\sqrt{x^2 + 9}} dx$

If $\displaystyle x = \frac{3}{sinh \theta}$, then $\displaystyle \frac{dx}{d\theta} = \frac{-3cosh \theta}{sinh^2 \theta}$.

Putting in the substitution, I get $\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{\frac{3}{sinh \theta}\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh^2 \theta} d\theta$.

From this I moved the $\displaystyle sinh \theta$ to the top of the first fraction and cancelled with one of the $\displaystyle sinh^2 \theta$ on the denominator of the second fraction to give me this.

$\displaystyle \int_{4}^{3\sqrt{3}} \frac{1}{3\sqrt{\frac{9}{sinh^2 \theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

I have no idea what to do next, would really appreciate some help?

Use the identity $\displaystyle \cosh^2 \theta = 1 + \sinh ^2 \theta$ in your square root. Note: Be careful with your limits.

3. Originally Posted by Danny
Thanks I nearly forgot them, the new limits are $\displaystyle \ln{\sqrt{3}}$ and $\displaystyle \ln{2}$ respectively I think?

Originally Posted by Danny
Use the identity $\displaystyle \cosh^2 \theta = 1 + \sinh ^2 \theta$ in your square root.
This would give me:

$\displaystyle \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\cosh^2 \theta - 1} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

Sorry but not sure where to go from here?

4. $\displaystyle \frac{1}{\sqrt{\frac{9}{\sinh ^2 \theta} + 9}} = \frac{1}{\sqrt{\frac{9 + 9 \sinh ^2 \theta}{\sinh ^2 \theta}}} = \frac{\sinh \theta}{3\sqrt{1 + \sinh ^2\theta}} = \frac{\sinh \theta}{3 \cosh \theta}$

5. Originally Posted by craig
This would give me:

$\displaystyle \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\cosh^2 \theta - 1} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

Sorry but not sure where to go from here?
Hi craig,

We have $\displaystyle \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\sinh^2\theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

$\displaystyle = \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{9\sqrt{\frac{1+\sinh^2\theta}{\sinh^2\the ta}}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

$\displaystyle =\frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{\sqrt{\coth^2\theta}}\times \frac{cosh \theta}{sinh \theta} d\theta$

Can you continue?

6. Originally Posted by Danny
$\displaystyle \frac{1}{\sqrt{\frac{9}{\sinh ^2 \theta} + 9}} = \frac{1}{\sqrt{\frac{9 + 9 \sinh ^2 \theta}{\sinh ^2 \theta}}} = \frac{\sinh \theta}{3\sqrt{1 + \sinh ^2\theta}} = \frac{\sinh \theta}{3 \cosh \theta}$
Thankyou!

Simplifies down to $\displaystyle \frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} 1$

$\displaystyle \left[\frac{-\theta}{3}\right]$

Putting in the limits you get

$\displaystyle \frac{1}{3}\left(\frac{1}{2}(\ln{4} - \ln{3}\right)$

Which gives you $\displaystyle \frac{1}{6}\ln{\frac{4}{3}}$

Thanks again for the help!

7. Originally Posted by Sean12345
Hi craig,

We have $\displaystyle \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{3\sqrt{\frac{9}{\sinh^2\theta} + 9}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

$\displaystyle = \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{9\sqrt{\frac{1+\sinh^2\theta}{\sinh^2\the ta}}} \times \frac{-3cosh \theta}{sinh \theta} d\theta$

$\displaystyle =\frac{-1}{3} \int_{\ln{2}}^{\ln{\sqrt{3}}} \frac{1}{\sqrt{\coth^2\theta}}\times \frac{cosh \theta}{sinh \theta} d\theta$

Can you continue?