Question

Match the following List I with List II

Answer 1

(P) $l={\int }_0^{\pi }x\left({\sin }^2\right(\sin x)+{\cos }^2(\cos x\left)\right)dx$
$l={\int }_0^{\pi }(\pi -x)\left({\sin }^2\right(\sin x)+{\cos }^2(\cos x\left)\right)dx$
$2l=x{\int }_0^{\pi }\left({\sin }^2\right(\sin x)+{\cos }^2(\cos x\left)\right)dx$
$2l=2x{\int }_0^{\pi /2}\left({\sin }^2\right(\sin x)+{\cos }^2(\cos x\left)\right)dx$
Again $l=\pi {\int }_0^{\pi /2}\left({\sin }^2\right(\cos x)+{\cos }^2(\sin x\left)\right)dx$
$2l=\pi {\int }_0^{\pi /2}2dx\Rightarrow l=\pi .\frac{\pi }{2}=\frac{{\pi }^2}{2}$
(Q) Let $f\left(x\right)=2\sin \sqrt{x}$
Then $x.f^{\prime }\left(x\right)=x.\frac{2\cos \sqrt{x}}{\sqrt{x}}=\sqrt{x}\cos \sqrt{x}$
$l={\int }_0^{{\pi }^2/16}\left(f\right(x)+x{f}^{\prime }(x\left)\right)dx=\left(xf\right(x){)}_0^{{\pi }^2/16}=\frac{{\pi }^2}{8\sqrt{2}}$
(R) ${\int }_{-\pi /4}^{\pi /4}In\left(\sqrt{1+{\sin }^{2}x}\right)dx={\int }_{-\pi /4}^{\pi /4}In|\sin x+\cos x|dx$
$={\int }_{-\pi /4}^{\pi /4}In\left|\sqrt{2}\sin (x-\frac{\pi }{4})\right|dx={\int }_0^{\pi /4}In|\sqrt{2}\sin t|dt=-\frac{\pi }{4}\times In2$
(S) $l={\int }_0^{\pi }\frac{x^3{\cos }^{4}x{\sin }^{2}x}{{\pi }^2-3\pi x+3x^2}dx$
$2l={\int }_0^{\pi }{\cos }^{4}x{\sin }^{2}xdx$
$l=\pi {\int }_0^{\pi /2}{\cos }^{4}x{\sin }^{2}xdx=\frac{{\pi }^2}{32}$