Question

Let $a_n={\int }_0^{\pi /2}\frac{1-cos2nx}{1-cos2x}dx,$ then the value of $\left|\begin{array}{ccc}\frac{\pi }{2}& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$ equals

• A. 0
• B. 1
• C. 3
• D. 5

Answer 1

The correct option is A 0

$a_n={\int }_0^{\pi /2}\frac{1-cos2nx}{1-cos2x}dx,$
$a_{n+2}={\int }_0^{\pi /2}\frac{1-cos2(n+2)x}{1-cos2x}dx$
Also,$a_{n+1}={\int }_0^{\pi /2}\frac{1-cos2(n+1)x}{1-cos2x}dx$
$a_n+a_{n+2}={\int }_0^{\pi /2}\frac{2-\left[\cos 2nx+cos2\right(n+2\left)x\right]}{1-cos2x}dx$
$={\int }_0^{\pi /2}\frac{2-\left[2\cos 2(n+1)x\cos 2x\right]}{1-\cos 2x}dx$
Now, $a_n+a_{n+2}-2a_{n+1}={\int }_0^{\pi /2}\frac{-2\cos 2(n+1)xcos2x+2cos2(n+1)x}{1-cos2x}dx$
$={\int }_0^{\pi /2}\frac{2\cos 2(n+1)x(1-cos2x)}{1-cos2x}dx$
$=2{\int }_0^{\pi /2}\cos 2(n+1)xdx$
$={\left[\frac{\sin 2(n+1)x}{2(n+1)}\right]}_0^{\pi /2}$
$=0$
Now, consider, $\left|\begin{array}{ccc}\frac{\pi }{2}& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$
$C_1\to C_1+C_3-2C_2$
$=\left|\begin{array}{ccc}0& a_2& a_3\\ 0& a_5& a_6\\ 0& a_8& a_9\end{array}\right|$
$=0$