# Solving an integral using ash substitution

#### 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: $$\displaystyle ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}$$

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

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

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

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

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

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

There must be an easier way yo solve this.

#### HallsofIvy

MHF Helper
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.

Ulysses

#### Ulysses

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)?

#### Prove It

MHF Helper
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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ulysses

#### HallsofIvy

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

Ulysses

#### Prove It

MHF Helper
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...