Question

The area bounded by the curves $y=si{n}^{-1}|sinx|$ and $y=(si{n}^{-1}|sinx|{)}^2,$ where $0\le x\le 2\pi ,$ is:

• A. $\frac{1}{3}+\frac{{\pi }^2}{4}$ sq. units
• B. $\frac{1}{6}+\frac{{\pi }^3}{8}$ sq. units
• C. $2$ sq. units
• D. $\frac{4}{3}+{\pi }^2\frac{\pi -3}{6}sq.units$

Answer 1

The correct option is D $\frac{4}{3}+{\pi }^2\frac{\pi -3}{6}sq.units$

$y=si{n}^{-1}\left|sinx\right|=\{\begin{array}{cc}x& 0\le x<\pi /2\\ \pi -x& \pi /2\le x<\pi \\ x-\pi & \pi \le x<3\pi /2\\ 2\pi -x& 3\pi /2\le x<2\pi \end{array}$
$y={\left(si{n}^{-1}\left|sinx\right|\right)}^2=\{\begin{array}{cc}{x}^2& 0\le x<\pi /2\\ {(\pi -x)}^2& \pi /2\le x<\pi \\ {(x-\pi )}^2& \pi \le x<3\pi /2\\ {(2\pi -x)}^2& 3\pi /2\le x<2\pi \end{array}$
Therefore required area $=4{\int }_0^1{(x-x^2)}^{2}dx+{\int }_1^{\frac{\pi }{2}}(x^2-x)dx$
$=\frac{4}{3}+{\pi }^2\frac{\pi -3}{6}sq.units$