Question

If, $I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n}xdx,n>1\text{and}n\in N,$ then which of the following option(s) is/are true

• A. $I_n+I_{n-2}=\frac{1}{n+1}$
• B. $I_n+I_{n-2}=\frac{1}{n-1}$
• C. $I_2+I_4,I_4+I_6,\cdots \text{are in H.P}$
• D. $\frac{1}{2(n+1)}<I_n<\frac{1}{2(n-1)}$

Answer 1

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

$\frac{1}{2(n+1)}<I_n<\frac{1}{2(n-1)}$

If $I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n}xdx$,
then $I_n+I_{n-2}=\frac{1}{n-1}$
Now putting $n=4$
$\Rightarrow I_2+I_4=\frac{1}{3}$
Putting $n=6$
$\Rightarrow I_6+I_4=\frac{1}{5}$ and so on
$\Rightarrow I_2+I_4,I_4+I_6$ are in H.P.
For, $0<x<\frac{\pi }{4}$ we have
$0<{\tan }^{n+2}x<{\tan }^{n}x<{\tan }^{n-2}x\Rightarrow I_{n+2}<I_n<I_{n-2}$
$\Rightarrow I_n+I_{n+2}<2I_n<I_n+I_{n-2}\Rightarrow \frac{1}{n+1}<2I_n<\frac{1}{n-1}\Rightarrow \frac{1}{2(n+1)}<I_n<\frac{1}{2(n-1)}$