Question

If $I_n={\int }_0^{\pi /4}{\tan }^{n}dx,$ then $\frac{1}{I_2+I_4}\frac{1}{I_3+I_5}\frac{1}{I_4+I_6}$ is :

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is A A.P.

$I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$
$\Rightarrow I_{n+2}+I_n={\int }_0^{\pi /4}{\tan }^{n+2}xdx+{\int }_0^{\pi /4}{\tan }^{n}xdx$
$={\int }_0^{\pi /4}{\tan }^{n}x(1+{\tan }^{2}x)dx={\int }_0^{\pi /4}{\tan }^{n}x{\sec }^{2}xdx.$
$={\int }_0^1{t}^{n}dt$ where $t=\tan x=\frac{1}{n+1}$
$\Rightarrow I_2+I-4=\frac{1}{3},I_3+I_5=\frac{1}{4},I_4+I_6=\frac{1}{5}$
$\Rightarrow \frac{1}{I_2+I_4}=3,\frac{1}{I_3+I_5}=4,\frac{1}{I_4+I_6}=5$
$\Rightarrow \frac{1}{I_2+I_4},\frac{1}{I_3+I_5},\frac{1}{I_4+I_6}ϵA.P.$