Question

If $I_n=\underset{−π}{\overset{π}{∫}}\frac{\sin nx}{\left(1+{π}^x\right)\sin x}dx,n=0,1,2,…$, then

• A. $I_n=I_{n+2}$
• B. $\underset{m=1}{\overset{10}{∑}}{I}_{2m+1}=10Ï€$
• C. $\underset{m=1}{\overset{10}{∑}}{I}_{2m}=10$
• D. $I_n=I_{n+1}$

Answer 1

Answer: (A), (B), (C)

Given $I_n=\underset{−π}{\overset{π}{∫}}\frac{\sin nx}{\left(1+{π}^x\right)\sin x}dx....$ (i)
Using $\underset{a}{\overset{b}{∫}}f(x)dx=\underset{a}{\overset{b}{∫}}f(b+a−x)dx$, we get
$I_n=\underset{−π}{\overset{π}{∫}}\frac{{π}^x\sin nx}{\left(1+{π}^x\right)\sin x}dx.....$(ii)
On adding Eqs. (i) and (ii), we have
$2I_n=\underset{−π}{\overset{π}{∫}}\frac{\sin nx}{\sin x}dx=2\underset{0}{\overset{π}{∫}}\frac{\sin nx}{\sin x}$
$[âˆ\mu f(x)=\frac{\sin n\sin x}{\sin }dx$ is an even function]
$⇒I_n=\underset{0}{\overset{π}{∫}}\frac{\sin nx}{\sin x}dx$
Now, $I_{n+2}−I_n=\underset{0}{\overset{π}{∫}}\frac{\sin (n+2)x−\sin nx}{\sin x}dx$
$=\underset{0}{\overset{Ï€}{∫}}\frac{2\cos (n+1)xâ‹\dots \sin x}{\sin x}dx$
$=2\underset{0}{\overset{π}{∫}}\cos (n+1)xdx=2{\left[\frac{\sin (n+1)x}{(n+1)}\right]}_0^{π}=0$
$∴I_{n+2}=I_n.....$(iii)
Since, $I_n=\underset{0}{\overset{π}{∫}}\frac{\sin nx}{\sin x}dx$
$⇒I_1=π$ and $I_2=0$
From Eq. (iii) $I_1=I_3=I_5=…=π$
and $I_2=I_4=I_6=…=0$
$⇒\underset{m=1}{\overset{10}{∑}}{I}_{2m+1}=10π$
and $\underset{m=1}{\overset{10}{∑}}{I}_{2m}=0$
$∴$ Correct options are (a), (b), (c).