Firstly,
$\displaystyle G(x) = \int_x^6 \cos(\sqrt{2t}) dt = - \int_6^x \cos(\sqrt{2t}) dt$
The FTC says it then follows that $\displaystyle G'(x) = -\cos(\sqrt{2x})$
Just in case a picture helps...
... and in this case...
... where straight continuous lines differentiate downwards (integrate up) with respect to x.
Spoiler:
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Don't integrate - balloontegrate!
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Balloon Calculus Drawing with LaTeX and Asymptote!