Question

If $I_n=\int {\sec }^{n}xdx,$ then $I_8-\frac{6I_6}{7}=$

• A. $-\frac{{\sec }^{6}x\cdot \tan x}{7}$
• B. $-\frac{\sec x\cdot {\tan }^{7}x}{7}$
• C. $\frac{{\sec }^{6}x\cdot \tan x}{7}$
• D. $\frac{\sec x\cdot \tan x}{7}$

Answer 1

The correct option is C $\frac{{\sec }^{6}x\cdot \tan x}{7}$

We know that,
$I_n=\int {\sec }^{n}xdx=\frac{\tan x\cdot {\sec }^{n-2}x}{n-1}+\frac{n-2}{n-1}{I}_{n-2}$
Putting $n=8,$ we get
$I_8=\int {\sec }^{8}xdx=\frac{\tan x\cdot {\sec }^{6}x}{7}+\frac{6}{7}{I}_6$
$\Rightarrow I_8-\frac{6}{7}{I}_6=\frac{\tan x\cdot {\sec }^{6}x}{7}$