Question

For $n\ge 2$, If $I_n=\int {\sec }^{n}xdx$, then $I_4-\frac{2}{3}{I}_2=$

• A. ${\sec }^{2}x\tan x+c$
• B. $\frac{1}{3}{\sec }^{2}x\tan x+c$
• C. $\frac{2}{3}{\sec }^{2}x\tan x+c$
• D. $\frac{1}{2}\log |\sec x+\tan x\mid +c$

Answer 1

Answer: (B)

We have,
$I_n=\int {\sec }^{n}xdx$
$\therefore I_2=\int {\sec }^{2}xdx=\tan x+c_1$
and $I_4=\int {\sec }^{4}xdx$
$=\int {\sec }^{2}x\cdot {\sec }^{2}xdx$
$=\int \left({\tan }^{2}x+1\right){\sec }^{2}xdx$
$=\frac{{\tan }^{3}x}{3}+\tan x+c_2$
$\therefore I_4-\frac{2}{3}{I}_2=\frac{{\tan }^{3}x}{3}+\tan x+c_2-\frac{2}{3}\tan x-\frac{2c_1}{3}$
$=\frac{1}{3}{\tan }^{3}x+\frac{1}{3}\tan x+c$
$[$ where $c_2-\frac{2}{3}{c}_1=c]$
$=\frac{1}{3}\tan x\left({\tan }^{2}x+1\right)+c$
$=\frac{1}{3}\tan x{\sec }^{2}x+c$