Question

The area bounded by the curve $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3\pi }{2}$ is

Answer 1

The desired are $={\int }_0^{\pi /4}\left(cosx\right)dx$

$+{\int }_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx+{\int }_{5\pi /4}^{3\pi /2}(\cos x-\sin x)dx$

$=2[\sin x+\cos x{]}_0^{\pi /4}+[-\cos x-\sin x{]}_{\pi /4}^{5\pi /4}$

(As the first and third integrals are equal in magnitude )

$=2(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)+(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$

$\frac{8}{\sqrt{2}}-2=4\sqrt{2}-2$