Question

For $n \geq 2$, If $I_{n}=\int \sec ^{n} x d x$, 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,
$I_{n}=\int \sec ^{n} x d x$
$\therefore I_{2}=\int \sec ^{2} x d x=\tan x+c_{1}$
and $I_{4}=\int \sec ^{4} x d x$
$=\int \sec ^{2} x \cdot \sec ^{2} x d x$
$=\int\left(\tan ^{2} x+1\right) \sec ^{2} x d x$
$=\frac{\tan ^{3} x}{3}+\tan x+c_{2}$
$\therefore I_{4}-\frac{2}{3} I_{2}=\frac{\tan ^{3} x}{3}+\tan x+c_{2}-\frac{2}{3} \tan x-\frac{2 c_{1}}{3}$
$=\frac{1}{3} \tan ^{3} x+\frac{1}{3} \tan x+c$
$\left[\right.$ where $\left.c_{2}-\frac{2}{3} c_{1}=c\right]$
$=\frac{1}{3} \tan x\left(\tan ^{2} x+1\right)+c$
$=\frac{1}{3} \tan x \sec ^{2} x+c$