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.