In class today, we had a problem in which we needed to integrate a function following the form

 

[math]

\int{\sqrt{u^2 \pm a^2}\,du}

[/math]

 

and it said to refer to the appendices which gave that integral was equal to

 

[math]

\frac{1}{2}(u\sqrt{u^2+a^2}\pm a^2\ln{|u+\sqrt{u^2+a^2}}|)

[/math].

 

I tried to do this (using [math]+[/math] for the [math]\pm[/math]) and I first did some rationalizing

 

[math]

\int{\sqrt{u^2 \pm a^2}} \frac{\sqrt{u^2 \pm a^2}}{\sqrt{u^2 \pm a^2}}\,du

[/math]

 

as to split it into

 

[math]

\int{\frac{u^2+a^2}{\sqrt{u^2+a^2}}du} = \int{\frac{u^2}{\sqrt{u^2+a^2}}du} + \int{\frac{a^2}{\sqrt{u^2+a^2}}du}

[/math]

 

and the second fraction is, if I understand correctly,

 

[math]

\int{\frac{a^2}{\sqrt{u^2+a^2}}du} = a^2 \sinh^{-1}\frac{u}{a} = a^2\ln{|u+\sqrt{u^2+a^2}|}

[/math]

 

which would reason that

 

[math]

\int{\frac{u^2}{\sqrt{u^2+a^2}}du} = \frac{1}{2}(u\sqrt{u^2+a^2}-a^2\ln{|u+\sqrt{u^2+a^2}|})

[/math]

 

but I can't find how to do this. Am I overlooking a simple technique? I also can't look up the proof because I don't know the name or whatever. Can anyone explain this last step, or did I start wrong?

 

(Either the whole derivation or just how to integrate [math]

\int{\frac{u^2}{\sqrt{u^2+a^2}}du} [/math] )

I don't think this is the approach you should be taking. The very last integral you state is quite non-obvious to integrate at first sight. However, you can solve your original equation by simply letting [math]u = a\sinh t[/math]. Then it transforms into:

 

[math]a^2 \int \cosh^2 t \, dt[/math]

 

by using [math]\cosh^2 t - \sinh^2 t = 1[/math]. Evaluating this integral is easy by using the double angle formulae for cosh² in terms of cosh(2x). The result follows from taking the inverse of your substitution.

I don't think this is the approach you should be taking. The very last integral you state is quite non-obvious to integrate at first sight. However, you can solve your original equation by simply letting [math]u = a\sinh t[/math]. Then it transforms into:

 

[math]a^2 \int \cosh^2 t \, dt[/math]

 

by using [math]\cosh^2 t - \sinh^2 t = 1[/math]. Evaluating this integral is easy by using the double angle formulae for cosh² in terms of cosh(2x). The result follows from taking the inverse of your substitution.

 

Oh thanks! I'm new to the hyperbolic functions so I didn't think that at first. It looks much simpler now.