JEE Main · 2010 · 11 Apr 2010 · Offline · Area Under Curves

Between $x=0$ and $x=\frac{3\pi }{2}$ the curves $y=\cos x$ and $y=\sin x$ cross where $\tan x=1$, namely at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$. Taking the upper curve minus the lower on each stretch, the area is ${\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx+{\int }_{5\pi /4}^{3\pi /2}(\cos x-\sin x)dx$. The three pieces evaluate to $(\sqrt{2}-1)$, $2\sqrt{2}$ and $(\sqrt{2}-1)$, whose sum is $4\sqrt{2}-2$.