Question

If $I={\int }_0^{\pi }\frac{x^2{\sin }^{2}x{\cos }^{4}x}{x^2-3\pi x+3x^2}dx$ then the value of $\frac{32}{{\pi }^2}I+298$ is equal

Answer 1

Given : $I={\int }_0^{\pi }\frac{x^3{\cos }^{4}x{\sin }^{2}xdx}{{\pi }^2-3\pi x+3x^2}dx...........\left(1\right)$

${\int }_0^{a}f\left(x\right)dx={\int }_b^{a}f(a-x)dxI={\int }_0^{\pi }\frac{(\pi -x{)}^3{\cos }^4(\pi -x){\sin }^2(\pi -x)}{({\pi }^2-3\pi (\pi -x)+3(\pi -x{)}^2)}I={\int }_0^{\pi }\frac{(\pi -x{)}^3{\cos }^{4}x{\sin }^{2}x}{({\pi }^2-3\pi x)+3x^2)}...........\left(2\right)$

Adding (1) and (2)

$2I={\int }_0^{\pi }\frac{(\pi -x{)}^3+x^3}{({\pi }^2-3\pi x+3x^2)}{\sin }^{2}x{\cos }^{4}xdx2I={\int }_0^{\pi }\frac{(x+\pi -x)\left[\right(\pi -x{)}^2+x^2-x(\pi -x)]{\sin }^{2}x{\cos }^{4}xdx}{({\pi }^2-3\pi x+3x^2)}2I={\int }_0^{\pi }\frac{\pi (x^2+{\pi }^2-2\pi x+x^2+\pi x+x^2){\sin }^{2}x{\cos }^{4}xdx}{{(\pi }^2-3\pi x+3x^2)}2I=\pi {\int }_0^{\pi }\frac{{(\pi }^2-3\pi x+3x^2){\sin }^{2}x{\cos }^{4}xdx}{{(\pi }^2-3\pi x+3x^2)}2I=\pi {\int }_0^{\pi }{\sin }^{2}x{\cos }^{2}xdx2I=\pi 2{\int }_0^{\frac{\pi }{2}}\left({\sin }^{2}x{\cos }^{4}x\right)dx............\left(3\right)2I=\pi 2{\int }_0^{\frac{\pi }{2}}\left({\sin }^{2}x{\cos }^{4}x\right)dx.................\left(4\right)$

Adding (3) and (4)

$4I=2\pi {\int }_0^{\frac{\pi }{2}}({\sin }^{2}x{\cos }^{4}x+{\cos }^{2}x{\sin }^{4}x)dx4I=2\pi {\int }_0^{\frac{\pi }{2}}\left({\sin }^{2}x{\cos }^{2}x\right)=\frac{2\pi }{4}{\int }_0^{\frac{\pi }{2}}\left({\sin }^{2}2x\right)dx4I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}{\sin }^{2}2xdx.........\left(5\right)4I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}{\cos }^{2}2xdx..........\left(6\right)$

Adding (5) and (6)

$8I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}dxI=\frac{{\pi }^2}{32}\frac{32}{{\pi }^2}I+298=299$

Hence the correct answer is $299$