Question

If $I_n={\int }_0^{\pi }\frac{1-\sin 2nx}{1-\cos 2x}dx$ 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

$I_n={\int }_0^{\pi }\frac{1-sin2nx}{1-cos2x}dx$
Consider, $I_n+I_{n+2}-2I_{n+1}={\int }_0^{\pi }\frac{1-sin2nx+(1-sin(2n+4\left)x\right)-2(1-sin(2n+2\left)x\right)}{1-cos2x}dx$
$={\int }_0^{\pi }\frac{2sin(2n+2)x-\left(sin\right(2n+4)x+sin2nx)}{1-cos2x}dx$
$={\int }_0^{\pi }\frac{2sin(2n+2)x-2sin(2n+2)xcos2x}{1-cos2x}dx$
$=2{\int }_0^{\pi }sin(2n+2)xdx$
$={\frac{-2cos(2n+2)x}{2n+2}}_0^{\pi }$
$=\frac{-1}{n+1}\left[cos\right(n+1)2\pi -cos0]$
$=\frac{-1}{n+1}(1-1)=0$
$\Rightarrow 2I_{n+1}=I_n+I_{n+2}$
$\therefore I_n,I_{n+1},I_{n+2},....$ are in A.P.