A) \[\frac{{{\pi }^{5}}}{5}\]
B) \[\frac{{{\pi }^{5}}}{10}\]
C) \[\frac{2{{\pi }^{5}}}{5}\]
D) None of these
Correct Answer: A
Solution :
[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, \[A=\int\limits_{0}^{2x}{\frac{{{(y-\pi )}^{4}}}{1+{{\tan }^{4}}\frac{y}{4}}}dy=\int\limits_{0}^{2\pi }{\frac{{{(2\pi -y-\pi )}^{4}}}{1+{{\tan }^{4}}\frac{2\pi -y}{4}}}dy\] \[=\int\limits_{0}^{2\pi }{\frac{{{(y-\pi )}^{4}}}{1+{{\cot }^{4}}\frac{y}{4}}}dy=\int\limits_{0}^{2\pi }{\frac{{{\tan }^{4}}\frac{y}{4}{{(y-\pi )}^{4}}}{{{\tan }^{4}}\frac{y}{4}+1}}dy\] \[\therefore \,\,\,\,\,\,\,\,\,2A=\int\limits_{0}^{2\pi }{{{(y-\pi )}^{4}}dy}=2\int\limits_{6}^{\pi }{{{(y-\pi )}^{4}}dy}\] \[=2\,\,\,\left[ \frac{{{(y-\pi )}^{5}}}{5} \right]_{0}^{\pi }=\frac{2{{\pi }^{5}}}{5}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,A=\frac{{{\pi }^{5}}}{5}\]
You need to login to perform this action.
You will be redirected in
3 sec
