Question

If $I_n={\int }_0^{\pi /2}\frac{{\cos }^{2}nx}{\sin x}dx$, then $I_2-I_1,I_3-I_2,I_4-I_3$ are in

• A. GP
• B. AP
• C. HP
• D. None of these

Answer 1

The correct option is C HP

We have,
$I_n-I_{n-1}={\int }_0^{\pi /2}\frac{{\cos }^{2}nx-{\cos }^2(n-1)x}{\sin x}dx$
$={\int }_0^{\pi /2}\frac{{\sin }^2(n-1)x{\sin }^{2}nx}{\sin x}dx$
$={\int }_0^{\pi /2}\sin (2n-1)xdx$
$=\frac{1}{2n-1}{\left[\cos (2n-1)x\right]}_0^{\pi /2}$
$=\frac{-1}{2n-1}$
$\therefore I_2-I_1=-\frac{1}{3}$
$I_3-I_2=-\frac{1}{5}$
$I_4-I_3=-\frac{1}{7}$
Clearly, $I_2-I_1,I_3-I_2,I_4-I_3$ are in HP