Solving an integral using ash substitution

May 2010
254
8
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: \(\displaystyle ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}\)


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

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

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

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

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


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

There must be an easier way yo solve this.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
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May 2010
254
8
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

HallsofIvy said:
sh and ch are the hyperbolic function, sinh(t) and cosh(t)?
The answer is yes :p
 

Prove It

MHF Helper
Aug 2008
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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: \(\displaystyle ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}\)


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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

MHF Helper
Apr 2005
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7,909
But notice in the original post, "The statement says that I must solve it using an adequate trigonometric substitution". (Emphasis mine.)
 
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Prove It

MHF Helper
Aug 2008
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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...