Question

If $I_n=\underset{0}{\overset{\pi }{\int }}\frac{1-\cos nx}{1-\cos x}dx,$ where $n$ is whole number, then

• A. $I_n,I_{n+1},I_{n+2}$ are in $G.P.$
• B. $I_n,I_{n+1},I_{n+2}$ are in $A.P.$
• C. $I_n=n\pi$
• D. $\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}n\theta }{{\sin }^2\theta }d\theta =\frac{n\pi }{2}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $I_n,I_{n+1},I_{n+2}$ are in $A.P.$
C $I_n=n\pi$
D $\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}n\theta }{{\sin }^2\theta }d\theta =\frac{n\pi }{2}$

$I_{n+1}-I_n=\underset{0}{\overset{\pi }{\int }}\frac{(1-\cos (n+1)x)-(1-\cos \left(nx\right))}{1-\cos x}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{\cos \left(nx\right)-\cos (n+1)x}{1-\cos x}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{2\sin (nx+\frac{x}{2})\sin \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$
$\Rightarrow I_{n+1}-I_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin (nx+\frac{x}{2})}{\sin \frac{x}{2}}dx\cdots \left(1\right)$

Putting $n\to n+1$
$I_{n+2}-I_{n+1}=\underset{0}{\overset{\pi }{\int }}\frac{\sin (nx+\frac{3x}{2})}{\sin \frac{x}{2}}dx\cdots \left(2\right)$

Subtraction $\left(1\right)$ from $\left(2\right)$,
$I_{n+2}+I_n-2I_{n+1}=\underset{0}{\overset{\pi }{\int }}\frac{\sin (nx+\frac{3x}{2})-\sin (nx+\frac{x}{2})}{\sin \frac{x}{2}}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{2\cos (n+1)x\cdot \sin \frac{x}{2}}{\sin \frac{x}{2}}dx$
$=2\underset{0}{\overset{\pi }{\int }}\cos (n+1)xdx=\frac{2}{n+1}[\sin (n+1)x{]}_0^{\pi }\Rightarrow I_{n+2}+I_n-2I_{n+1}=0$
Therefore, $I_n,I_{n+1},I_{n+2}$ are in $A.P.$

Now,
$I_0=0I_1=\pi$
So, the common difference is $\pi$
$I_n=I_0+(n+1-1)\pi =n\pi$

Let
$U=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}n\theta }{{\sin }^2\theta }d\theta \Rightarrow U=\underset{0}{\overset{\pi /2}{\int }}\frac{1-\cos 2n\theta }{1-\cos 2\theta }d\theta$
Putting $2\theta =x\Rightarrow 2d\theta =dx$
$\Rightarrow U=\frac{1}{2}\underset{0}{\overset{\pi }{\int }}\frac{1-\cos nx}{1-\cos x}dx=\frac{n\pi }{2}$

Alternate Solution:
$I_n=\underset{0}{\overset{\pi }{\int }}\frac{1-\cos nx}{1-\cos x}dx$

$I_0=0I_1=\underset{0}{\overset{\pi }{\int }}1dx=\pi$
$I_2=\underset{0}{\overset{\pi }{\int }}\frac{1-\cos 2x}{1-\cos x}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2}x}{{\sin }^2\left(x/2\right)}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{4{\sin }^2\left(x/2\right){\cos }^2\left(x/2\right)}{{\sin }^2\left(x/2\right)}dx$
$=\underset{0}{\overset{\pi }{\int }}2(1+\cos x)dx$
$=2[x+\sin x{]}_0^{\pi }$
$=2\pi$

So, the common difference is $\pi .$
$\therefore I_n=I_0+(n+1-1)\pi =n\pi$

Let
$U=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}n\theta }{{\sin }^2\theta }d\theta \Rightarrow U=\underset{0}{\overset{\pi /2}{\int }}\frac{1-\cos 2n\theta }{1-\cos 2\theta }d\theta$
Putting $2\theta =x\Rightarrow 2d\theta =dx$
$\Rightarrow U=\frac{1}{2}\underset{0}{\overset{\pi }{\int }}\frac{1-\cos nx}{1-\cos x}dx=\frac{1}{2}{I}_n=\frac{n\pi }{2}$