Question

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

• A. an A.P
• B. a G.P
• C. a H.P
• D. an AGP

Answer 1

The correct option is A an A.P

$I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$

$I_{n+2}={\int }_0^{\pi /4}{\tan }^{n}x\left({\tan }^{2}x\right)dx$

$={\int }_0^{\pi /4}{\tan }^{n}x({\sec }^{2}x-1)dx$

$={\int }_0^{\pi /4}{\tan }^{n}xd\left(\tan x\right)-I_n$

$\Rightarrow I_{n+2}+I_n=\frac{1}{n+1}$

$I_2+I_4=\frac{1}{3}$

$I_3+I_5=\frac{1}{4}$

$I_4+I_6=\frac{1}{5}$

$So,\frac{1}{I_2+I_4};\frac{1}{I_3+I_5};\frac{1}{I_4+I_6}$ forms an AP