Question

Find the area of one of the curvilinear triangles formed by $y=\sin x,y=\cos x$ and $x$ axis

Answer 1

The point of intersection of two curves $y=\sin x$ and $y=\cos x$ is at $(\frac{\pi }{4},\frac{1}{\sqrt{2}})$

Required Area$={\int }_0^{\frac{\pi }{4}}\sin xdx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cos xdx$

$={[-\cos x]}_0^{\frac{\pi }{4}}+{\left[\sin x\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=-[1-\frac{1}{\sqrt{2}}]+[\frac{1}{\sqrt{2}}-1]$

$=2[\frac{1}{\sqrt{2}}-1]$

$=\frac{2}{\sqrt{2}}-2$

$=\frac{2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}-2$

$=\sqrt{2}-2$sq.unit