Question

If $I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x{\sec }^{2}xdx,$ then $I_1,I_2,I_3..$ are in

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

Answer 1

Answer: (C)

H.P.

$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x{\sec }^{2}xdx,$

Substituting $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

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

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

Reciprocal of the terms $I_1,I_2,I_3..$ are in A.P, therefore $\frac{1}{n+1},\frac{1}{n+2},\frac{1}{n+3}$ are in H.P.

Hence, option 'C' is correct.