Question

If $\int e^{\sec x}\left(\sec x\tan xf\right(x)+(\sec x\tan x+{\sec }^{2}x\left)\right)dx=e^{\sec x}f\left(x\right)+C$, then a possible choice of $f\left(x\right)$ is :

• A. $\sec x-\tan x-\frac{1}{2}$
• B. $\sec x+\tan x+\frac{1}{2}$
• C. $x\sec x+\tan x+\frac{1}{2}$
• D. $\sec x+x\tan x-\frac{1}{2}$

Answer 1

The correct option is B $\sec x+\tan x+\frac{1}{2}$

$\int e^{\sec x}\left(\sec x\tan xf\right(x)+(\sec x\tan x+{\sec }^{2}x\left)\right)dx=e^{\sec x}f\left(x\right)+C$
Differentiating both sides, w.r.t. $x$, we get
$e^{\sec x}\left(\sec x\tan xf\right(x)+(\sec x\tan x+{\sec }^{2}x\left)\right)=e^{\sec x}\cdot \sec x\tan xf\left(x\right)+e^{\sec x}{f}^{\prime }\left(x\right)$

$\therefore f^{\prime }\left(x\right)=\sec x\tan x+{\sec }^{2}x$
$\Rightarrow f\left(x\right)=\sec x+\tan x+c$