Question

If $I_n={\int }_0^{\pi /4}{\tan }^n\left(x\right)dx$ then $I_2+I_4,+I_3+I_5,+I_4+I_6,...$ are in ?

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. $A.G.P$

Answer 1

Answer: (C)

$H.P$

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

$I_n+I_{n+2}={\int }_0^{\pi /4}({\tan }^n\left(x\right)+{\tan }^{n+2}\left(x\right))dx={\int }_0^{\pi /4}{\tan }^n\left(x\right)(1+{\tan }^{2}x)dx={\int }_0^{\pi /4}{\tan }^n\left(x\right){\sec }^{2}xdx$

Let $\tan x=t$

${\sec }^{2}xdx=dt$

${\int }_0^{\pi /4}{\tan }^n\left(x\right){\sec }^{2}xdx={\int }_0^1{t}^{n}dt$

${\int }_0^1{t}^{n}dt=\frac{1}{n+1}$

So, the terms are 1/3,1/4,1/5...(H.P)