Question

If $I=\int se{c}^{6}xdx=\frac{1}{5}ta{n}^{3}x+Ata{n}^{3}x+tanx+C$ then A is equal to

• A. $1/3$
• B. $2/3$
• C. $-1/3$
• D. $-2/3$

Answer 1

The correct option is B $2/3$

Let $I=\int {\sec }^{6}xdx$
Using $\int {\sec }^{m}xdx=\frac{\sin x{\sec }^{m-1}x}{m-1}+\frac{m-2}{m-1}\int {\sec }^{-2+m}xdx$
$I=\frac{1}{5}\tan x{\sec }^{4}x+\frac{4}{5}\int {\sec }^{4}xdx$
$=\frac{1}{5}\tan x{\sec }^{4}x+\frac{4}{15}\tan x{\sec }^{2}x+\frac{8}{15}\int {\sec }^{2}xdx$
$=\frac{1}{5}\tan x{\sec }^{4}x+\frac{4}{15}\tan x{\sec }^{2}x+\frac{8}{15}\tan x+c$
$=\frac{1}{5}\tan x(1+{\tan }^{2}x)(1+{\tan }^{2}x)+\frac{4}{15}\tan x(1+{\tan }^{2}x)+\frac{8}{15}\tan x+c$
$=\frac{1}{5}{\tan }^{5}x+\frac{2}{3}{\tan }^{3}x+\tan x+c$
Comparing the above equation with $\int ta{n}^{6}xdx=\frac{1}{5}ta{n}^{5}x+Ata{n}^{3}x+tanx-x+C$

we get, $A=\frac{2}{3}$