Question

If $\int {\sec }^{6}xdx=\frac{1}{5}{\tan }^{5}x+A{\tan }^{3}x+\tan x+c$, then $A$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

$\int {\sec }^{6}xdx$
$=\int {\sec }^{4}x.{\sec }^{2}xdx$

$=\int (1+{\tan }^{2}x{)}^2{\sec }^{2}x.dx$
$=\int (1+{\tan }^{4}x+2{\tan }^{2}x){\sec }^{2}x.dx$

$=\int {\sec }^{2}xdx+\int {\tan }^{4}x{\sec }^{2}x.dx+2\int {\tan }^{2}x{\sec }^{2}x.dx$

Put $\tan x=t$, so ${\sec }^{2}xdx=dt$
$=\int dt+\int t^{4}dt+\int 2t^{2}dt$

$=t+\frac{t^5}{5}+2\frac{t^3}{3}+c$

$=\tan x+\frac{{\tan }^{5}x}{5}+\frac{2}{3}{\tan }^{3}x+c.$
So, $A=\frac{2}{3}$