Question

If $I_n={\int }_0^{\pi /2}\frac{{\sin }^{2}nx}{{\sin }^{2}x}dx,$ then $I_1,I_2,I_3,$ are in :

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is A A.P.

Given $I_n={\int }_0^{\pi /2}\frac{{\sin }^{2}nx}{{\sin }^{2}x}dx$
Then $I_{n+1}-2I_n+I_{n-1}$
$={\int }_0^{\pi /2}\frac{{\sin }^2(n+1)x-2{\sin }^{2}nx+{\sin }^2(n-1)x}{{\sin }^{2}x}dx$
$={\int }_0^{\pi /2}\frac{\sin (2n+1)x-\sin x+{\sin }^2(n-1)x}{{\sin }^{2}x}dx$
$={\int }_0^{\pi /2}\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx$
$={\int }_0^{\pi /2}\frac{2\sin x\cos x}{\sin x}dx=2{\int }_0^{\pi /2}\cos 2nxdx$
$=\frac{1}{n}[\sin 2nx{]}_0^{\pi /2}=\frac{1}{n}[\sin n\pi -0]=0$
$\therefore I_{n+1}+I_{n-1}=2I_n\forall n\ge 1\Rightarrow I_1,I_2,I_3ϵA.P.$