# Thread: Solving an integral using ash substitution

1. ## Solving an integral using ash substitution

Hi there, I must solve this:

$\displaystyle \displaystyle\int_{}^{}x^2\sqrt[ ]{x^2+3}dx$

The statement says that I must solve it using an adequate trigonometric substitution.

I followed this way:

$\displaystyle x=\sqrt[ ]{3}sh(t)$
$\displaystyle dx=\sqrt[ ]{3}ch(t)dt$

I've used the identity: $\displaystyle ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}$

$\displaystyle u=ch(t)$
$\displaystyle du=sh(t)dt$

$\displaystyle dv=ch(t)dt$
$\displaystyle v=sh(t)$

$\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh(t)sh(t)dt$

$\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}(ch^2(t)-1)dt$

$\displaystyle *$ $\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=\displaystyle\fra c{1}{2}(ch(t)sh(t)+t)$

$\displaystyle 9\displaystyle\int_{}^{}sh^2(t)ch^2(t)dt=9(\displa ystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}(ch(t)sh(t)+t)sh(t)ch(t)dt= 9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}ch(t)^2sh(t)^2dt-\displaystyle\int_{}^{}ch(t)sh(t)tdt$

There must be an easier way yo solve this.

2. sh and ch are the hyperbolic function, sinh(t) and cosh(t)?

One other method would be to use trig functions rather than hyperbolic functions. Since $\displaystyle sin^2(t)+ cos^2(t)=1$, dividing both sides by $\displaystyle cos^2(t)$, $\displaystyle tan^2(t)+ 1= sec^2(t)$. That suggests using the substitution $\displaystyle x= \sqrt{3}tan(t)$. I doubt that that is any simpler but most people are more familiar with trig functions than with hyperbolic functions.

3. Thanks HallsofIvy. I thought of that too, but it was easier to memorize the hyperbolic substitutions to me, and I did all those kind of integrations using hyperbolic substitutions when I could, and $\displaystyle a\sin(t)=x$ in the others. I think you're right, but I'm not much familiarized with trigonometric identities, unless not much more than what I am with hyperbolic identities. But I'll try that way.

Bye there.

Oh, BTW

Originally Posted by HallsofIvy
sh and ch are the hyperbolic function, sinh(t) and cosh(t)?

4. Originally Posted by Ulysses
Hi there, I must solve this:

$\displaystyle \displaystyle\int_{}^{}x^2\sqrt[ ]{x^2+3}dx$

The statement says that I must solve it using an adequate trigonometric substitution.

I followed this way:

$\displaystyle x=\sqrt[ ]{3}sh(t)$
$\displaystyle dx=\sqrt[ ]{3}ch(t)dt$

I've used the identity: $\displaystyle ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}$

$\displaystyle u=ch(t)$
$\displaystyle du=sh(t)dt$

$\displaystyle dv=ch(t)dt$
$\displaystyle v=sh(t)$

$\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh(t)sh(t)dt$

$\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}(ch^2(t)-1)dt$

$\displaystyle *$ $\displaystyle \displaystyle\int_{}^{}ch^2(t)dt=\displaystyle\fra c{1}{2}(ch(t)sh(t)+t)$

$\displaystyle 9\displaystyle\int_{}^{}sh^2(t)ch^2(t)dt=9(\displa ystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}(ch(t)sh(t)+t)sh(t)ch(t)dt= 9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}ch(t)^2sh(t)^2dt-\displaystyle\int_{}^{}ch(t)sh(t)tdt$

There must be an easier way yo solve this.
$\displaystyle \int{x^2\sqrt{x^2 + 3}\,dx}$.

Let $\displaystyle x = \sqrt{3}\sinh{t}$ so that $\displaystyle dx = \sqrt{3}\cosh{t}\,dt$.

The integral becomes

$\displaystyle \int{(\sqrt{3}\sinh{t})^2\sqrt{(\sqrt{3}\sinh{t})^ 2 + 3}\,\sqrt{3}\cosh{t}\,dt}$

$\displaystyle = \int{3\sinh^2{t}\sqrt{3\sinh^2{t} + 3}\,\sqrt{3}\cosh{t}\,dt}$

$\displaystyle = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3(\sinh^2{t} + 1)}\,dt}$

$\displaystyle = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3\cosh^2{t}} \,dt}$

$\displaystyle = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3}\cosh{t}\, dt}$

$\displaystyle = \int{9\cosh^2{t}\sinh^2{t}\,dt}$

$\displaystyle = 9\int{\left(\frac{1}{2}\cosh{2t} + \frac{1}{2}\right)\left(\frac{1}{2}\cosh{2t} - \frac{1}{2}\right)\,dt}$

$\displaystyle = 9\int{\frac{1}{4}\cosh^2{2t} - \frac{1}{4}\,dt}$

$\displaystyle = \frac{9}{4}\int{\cosh^2{2t} - 1\,dt}$

$\displaystyle = \frac{9}{4}\int{\sinh^2{2t}\,dt}$

$\displaystyle = \frac{9}{4}\int{\frac{1}{2}\cosh{4t} - \frac{1}{2}\,dt}$

$\displaystyle = \frac{9}{8}\int{\cosh{4t} - 1\,dt}$

$\displaystyle = \frac{9}{8}\left[\frac{1}{4}\sinh{4t} - t\right] + C$

$\displaystyle = \frac{9}{32}\sinh{4t} - \frac{9}{8}t + C$

$\displaystyle = \frac{9}{32}(2\sinh{2t}\cosh{2t}) - \frac{9}{8}t + C$

$\displaystyle = \frac{9}{16}(\sinh{2t}\cosh{2t}) - \frac{9}{8}t + C$

$\displaystyle = \frac{9}{16}[2\sinh{t}\cosh{t}(2\sinh^2{t} + 1)] - \frac{9}{8}t + C$

$\displaystyle = \frac{9}{8}[\sinh{t}\sqrt{\sinh^2{t} + 1}(2\sinh^2{t} + 1)] - \frac{9}{8}t + C$

$\displaystyle = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\sqrt{\frac{x^2}{3} + 1}\left(\frac{2x^2}{3} + 1\right)\right] - \frac{9}{8}\sinh^{-1}\left(\frac{x}{\sqrt{3}}\right) + C$

$\displaystyle = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\sqrt{\frac{x^2 + 3}{3}}\left(\frac{2x^2 + 3}{3}\right)\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C$

$\displaystyle = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\frac{\sqrt{x^2 + 3}}{\sqrt{3}}\left(\frac{2x^2 + 3}{3}\right)\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C$

$\displaystyle = \frac{9}{8}\left[\frac{x(2x^2 + 3)\sqrt{x^2 + 3}}{9}\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} +C$

$\displaystyle = \frac{x(2x^2 + 3)\sqrt{x^2 + 3} - 9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C$

5. But notice in the original post, "The statement says that I must solve it using an adequate trigonometric substitution". (Emphasis mine.)

6. Originally Posted by HallsofIvy
But notice in the original post, "The statement says that I must solve it using an adequate trigonometric substitution". (Emphasis mine.)
Yet the title of the thread says "using ash substitution", in other words, a "sinh" subsitution...