Question

$\int {\sec }^{3}xdx=\frac{1}{2}[\sec x\tan x+\log (?+\tan x)]+c$
Choose the appropriate option to replace the question mark in the above equation.

• A. $\sin x$
• B. $\cos x$
• C. $\sec x$
• D. $\cot x$

Answer 1

Answer: (C)

$\sec x$

$I=\int {\sec }^{3}xdxd$
$=\int \sec x{\sec }^{2}xdx$
$=\int \sqrt{1+{\tan }^{2}x}.{\sec }^{2}xdx$
Put $\tan x=zor{\sec }^{2}xdx=dz$
$\therefore I=\int \sqrt{1+z^2}dz$
$=\frac{z\sqrt{z^2+1}}{2}+\frac{1}{2}\log |z+\sqrt{z^2+1}|+c$
$=\frac{\tan x\sec x}{2}+\frac{1}{2}\log (\tan x+\sec x)+c$
$=\frac{1}{2}[\sec x\tan x+\log (\sec x+\tan x)]+c$