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Check the picture below.
so we're looking for the area enclosed like so in the picture, now, where sin(x) and cos(x) meet or intersect at least on the 1st Quadrant will be at π/4 where both are 1/√2, so we're really looking for
[tex]\displaystyle\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}~~[sin(x)~~ - ~~cos(x)]dx\implies \int_{\frac{\pi }{4}}^{\frac{\pi }{2}}sin(x)dx~~ - ~~\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}cos(x)dx \\\\\\ \left. \cfrac{}{}-cos(x) \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}}~~ - ~~\left. \cfrac{}{}sin(x) \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}}\implies [0]~~ - ~~\left[-\cfrac{1}{\sqrt{2}} \right]~~ + ~~[-1]~~ - ~~\left[ -\cfrac{1}{\sqrt{2}} \right][/tex]
[tex]\cfrac{1}{\sqrt{2}}-1+\cfrac{1}{\sqrt{2}}\implies \cfrac{2}{\sqrt{2}}-1\implies \sqrt{2}-1[/tex]
