• # question_answer The area bounded by the curve $2{{(y-\pi )}^{4}}{{\cos }^{4}}\frac{y}{4}+x{{\sin }^{2}}\frac{y}{2}=2x$ and y-axis between $y=0$and $y=2\pi$ is A) $\frac{{{\pi }^{5}}}{5}$         B)                    $\frac{{{\pi }^{5}}}{10}$                    C) $\frac{2{{\pi }^{5}}}{5}$                   D)        None of these

[a]  We have             $x\left( 1-2{{\sin }^{2}}\frac{y}{4}{{\cos }^{2}}\frac{y}{4} \right)={{(y-\pi )}^{4}}{{\cos }^{4}}\frac{y}{4}$ $\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{{(y-\pi )}^{4}}{{\cos }^{4}}\frac{y}{4}}{1-2{{\sin }^{2}}\frac{y}{2}{{\cos }^{2}}\frac{y}{4}}=\frac{{{(y-\pi )}^{4}}}{{{\sec }^{4}}\frac{y}{4}-2{{\tan }^{2}}\frac{y}{2}}$ $=\frac{{{(y-\pi )}^{4}}}{\left( 1+{{\tan }^{2}}\frac{y}{4} \right)-2{{\tan }^{2}}\frac{y}{2}}=\frac{{{(y-\pi )}^{4}}}{1+{{\tan }^{4}}\frac{y}{4}}\ge 0$ $\therefore$  Required area, $A=\int\limits_{0}^{2x}{\frac{{{(y-\pi )}^{4}}}{1+{{\tan }^{4}}\frac{y}{4}}}dy=\int\limits_{0}^{2\pi }{\frac{{{(2\pi -y-\pi )}^{4}}}{1+{{\tan }^{4}}\frac{2\pi -y}{4}}}dy$ $=\int\limits_{0}^{2\pi }{\frac{{{(y-\pi )}^{4}}}{1+{{\cot }^{4}}\frac{y}{4}}}dy=\int\limits_{0}^{2\pi }{\frac{{{\tan }^{4}}\frac{y}{4}{{(y-\pi )}^{4}}}{{{\tan }^{4}}\frac{y}{4}+1}}dy$ $\therefore \,\,\,\,\,\,\,\,\,2A=\int\limits_{0}^{2\pi }{{{(y-\pi )}^{4}}dy}=2\int\limits_{6}^{\pi }{{{(y-\pi )}^{4}}dy}$ $=2\,\,\,\left[ \frac{{{(y-\pi )}^{5}}}{5} \right]_{0}^{\pi }=\frac{2{{\pi }^{5}}}{5}$ $\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,A=\frac{{{\pi }^{5}}}{5}$