Ok so I tried U-subbing sqrt(x^2-4), but I didn't get anywhere. Then I though hmm, maybe trig-sub will do the trick, but I also got stuck there. I don't know what's left to do!
edit:
The integral is:
sqrt(x-sqrt(x^2-4))
Ok so I tried U-subbing sqrt(x^2-4), but I didn't get anywhere. Then I though hmm, maybe trig-sub will do the trick, but I also got stuck there. I don't know what's left to do!
edit:
The integral is:
sqrt(x-sqrt(x^2-4))
Hyperbolic substitution will work... You should know that $\displaystyle \displaystyle \begin{align*} \cosh^2{t} - \sinh^2{t} \equiv 1 \implies \cosh^2{t} - 1 &\equiv \sinh^2{t} \end{align*}$, so the substitution $\displaystyle \displaystyle \begin{align*} x = 2\cosh{t} \implies dx = 2\sinh{t}\,dt \end{align*}$ will work.
$\displaystyle \displaystyle \begin{align*} \int{\sqrt{x^2 - 4}\,dx} &= \int{\sqrt{\left(2\cosh{t}\right)^2 - 4}\cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\cosh^2{t} - 4}\cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\left(\cosh^2{t} - 1\right)} \cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\sinh^2{t}}\cdot 2\sinh{t}\,dt} \\ &= \int{2\sinh{t} \cdot 2\sinh{t} \,dt} \\ &= 4\int{\sin^2{t}\,dt} \\ &= 4\int{\frac{1}{2}\cosh{2t} - \frac{1}{2}\,dt} \\ &= 4\left(\frac{1}{4}\sinh{2t} - \frac{1}{2}t\right) + C \\ &= \sinh{2t} - 2t + C \\ &= 2\sinh{t}\cosh{t} - 2t + C \\ &= 2\cosh{t}\sqrt{\cosh^2{t} - 1} - 2t + C \\ &= x\sqrt{\left(\frac{x}{2}\right)^2 - 1} - \textrm{arcosh}\,{\frac{x}{2}} + C \\ &= x\sqrt{\frac{x^2 - 4}{4}} - \textrm{arcosh}\,{\frac{x}{2}} + C \\ &= \frac{x\sqrt{x^2 - 4}}{2} - \textrm{arcosh}\,{\frac{x}{2}} + C \end{align*}$
what makes you think this integral has a closed-form antiderivative in the first place?
Wolfram's result ...
integrate sqrt(x - sqrt(4-x^2)) dx - Wolfram|Alpha
Wolfram DOES give me an answer...
integral[Sqrt[x - Sqrt[x^2 - 4]]] - Wolfram|Alpha
But no steps...
A shame it can't remember, but what Wolfram probably did is notice that the derivative of the integrand is virtually that integrand divided by
$\displaystyle \sqrt{x^2 - 4}$...
... where (key in spoiler) ...
Spoiler:
... and therefore do integration by parts (twice) on the product, $\displaystyle \frac{\sqrt{x - \sqrt{x^2 - 4}}}{\sqrt{x^2 - 4}}\ \sqrt{x^2 - 4}$ ...
... (key in spoiler) ...
Spoiler:
The rest...
Spoiler:
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