Solving an integral using a*sinh substitution

[Telemachus]

Homework Statement

The statement says: Calculate the next integrals using the adequate trigonometric substitution:

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

Homework Equations

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

The Attempt at a Solution

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

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

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

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

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

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

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

Well, if you see an easier way of solving this let me know :P

Bye there!

 

[vela]

You might want to try the substitution $x=\sqrt{3}\tan \theta$ instead.

 

[vela]

Or you could use your substitution and use the identities

$$\sinh 2x = 2\sinh x\cosh x$$

$$\sinh^2 x = \frac{\cosh 2x-1}{2}$$

instead of integration by parts.

 

[Telemachus]

You might want to try the substitution $x=\sqrt{3}\tan \theta$ instead.

Thanks vela. I'll try both ways.