JustStuit Posted April 3, 2007 Share Posted April 3, 2007 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] ) Link to comment Share on other sites More sharing options...

Dave Posted April 4, 2007 Share Posted April 4, 2007 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 cosh^{2} in terms of cosh(2x). The result follows from taking the inverse of your substitution. Link to comment Share on other sites More sharing options...

JustStuit Posted April 4, 2007 Author Share Posted April 4, 2007 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 cosh^{2} 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. Link to comment Share on other sites More sharing options...

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