Question

If $A_n={\int }_0^{\pi /2}\frac{\sin (2n-1)x}{\sin x}dx;B_n={\int }_0^{\pi /2}{\left(\frac{\sin nx}{\sin x}\right)}^{2}dx;$ for $n\in N$, then

• A. $A_{n+1}=A_n$
• B. $B_{n+1}=B_n$
• C. $A_{n+1}-A_n=B_{n+1}$
• D. $B_{n+1}-B_n=A_{n+1}$

Answer 1

Answer: (A), (D)

$A_{n+1}=A_n$
$B_{n+1}-B_n=A_{n+1}$

We have, $A_{n+1}-A_n={\int }_0^{\pi /2}\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx$

$={\int }_0^{\pi /2}\frac{2\cos 2nx\sin x}{\sin x}dx=2{\int }_0^{\pi /2}\cos 2nxdx=0$

Again $B_{n+1}-B_n={\int }_0^{\pi /2}\frac{{\sin }^2(n+1)x-{\sin }^{2}nx}{{\sin }^{2}x}dx$

$={\int }_0^{\pi /2}\frac{\sin (2n+1)x\sin x}{{\sin }^{2}x}dx=A_{n+1}$