Solving an integral using ash substitution

**Ulysses** · June 29th 2010, 03:42 PM

Hi there, I must solve this:

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

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

I followed this way:

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

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

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

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

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

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

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

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

There must be an easier way yo solve this.

**HallsofIvy** · June 29th 2010, 05:03 PM

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

**Ulysses** · June 29th 2010, 05:16 PM

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 $a\sin(t)=x$ 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)?

The answer is yes :P

**Prove It** · June 29th 2010, 06:42 PM

Originally Posted by Ulysses

Hi there, I must solve this:

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

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

I followed this way:

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

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

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

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

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

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

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

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

There must be an easier way yo solve this.

$\int{x^2\sqrt{x^2 + 3}\,dx}$ .

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

The integral becomes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**HallsofIvy** · June 30th 2010, 05:18 AM

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

**Prove It** · June 30th 2010, 08:09 PM

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...

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