$u=\cos^2(x) \implies du=-2\cos(x)\sin(x)dx \implies dx=\frac{du}{-2\cos(x)\sin(x)}$
Hieruit volgt:
\[\sin^2(x)=1-u \quad \text{en} \quad \sin(2x)^2=(2\sin(x)\cos(x))^2=4\sin(x)^2\cos(x)^2=4(1-u)u=4u-4u^2\]
De integraal wordt:
\[\int_{1}^{0}\frac{2\sin(x)\cos(x)}{\sqrt{4+2(1-u)-4u-\frac{4u-4u^2}{2}}}\frac{du}{-2\cos(x)\sin(x)}\]
\[=\int_{0}^{1}\frac{1}{\sqrt{4+2(1-u)-4u-\frac{4u-4u^2}{2}}}du\]
(Zie dat de volgorde van de grenzen omwisselt vanwege de min.)
Arccosh methode
\[\int_{0}^{1}\frac{1}{\sqrt{2u^2-8u+6}}du\] \[=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{1}{\sqrt{u^2-4u+3}}du\] \[=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{1}{\sqrt{u^2-4u+4-1}}du\] \[=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{1}{\sqrt{(u-2)^2-1}}du\] \[v=u-2 \implies dv=du\] \[\frac{1}{\sqrt{2}}\int_{-2}^{-1}\frac{1}{\sqrt{v^2-1}}dv\]
We gebruiken vervolgens de negatieve (omdat $v<\leq-1$) inverse hyberbolische cosinus:
\[=\frac{1}{\sqrt{2}}(-\operatorname{arccosh}(1)--\operatorname{arccosh}(2))\]
\[=\frac{\operatorname{arccosh}(2)}{2}\]