Question

If $I_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}co{s}^{n}xcosnxdx,then\sqrt{\frac{I_4}{I_8}}$ is equal to

Answer 1

$I_n=I_n=\underset{0}{\overset{\frac{\pi }{0}}{\int }}{\cos }^{n}x\cos nxdx$
$I_n=[{\cos }^{n}x\frac{\sin nx}{n}{]}_0^{\frac{\pi }{2}}-\underset{0}{\overset{\frac{\pi }{2}}{\int }}n{\cos }^{n-1}x(-\sin x).\frac{\sin nx}{n}dx$
$I_n=0+\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n-1}x\sin x\sin nxdx$
$\cos (n-1)x=\cos (nx-x)=\cos nx\cos x+\sin nx\sin x$
$I_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n-1}x\left(\cos \right(n-1)x-\cos nx\cos x)dx$
$I_n=I_{(n-1)}-I_n$
$\Rightarrow 2I_n=I_{n-1}$
$\therefore \frac{I_2}{I_1}=\frac{1}{2},\frac{I_3}{I_2}=\frac{1}{2},\frac{I_4}{I_3}=\frac{1}{2},\frac{I_5}{I_4}=\frac{1}{2},....\frac{I_8}{I_7}=\frac{1}{2}$
$\frac{I_8}{I_4}=\frac{1}{16}\Rightarrow \sqrt{\frac{I_4}{I_8}}=4$