Question

If $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx,$ then $\frac{1}{I_2+I_4},\frac{1}{I_3+I_5},\frac{1}{I_4+I_6},....$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none

Answer 1

The correct option is C H.P.

$I_n={\int }_0^{\pi /4}{\tan }^{n-2}x{\tan }^{2}xdx$

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

$I={\int }_0^{\pi /4}{\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\pi /4}{\tan }^{n-2}xdx$

$I_n={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\pi /4}-I_{n-2}$

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

Putting $n=4,5,6,....$ we get

$I_4+I_2,I_5+I_3,I_6+I_4,....$ are respectively $\frac{1}{3},\frac{1}{4},\frac{1}{5}\cdots$ which are in H.P. and hence their reciprocals are in A.P.