Question

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

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

Answer 1

Answer: (D)

$\int e^{\sec x}\left(\sec x\tan xf\left(x\right)+\left(\sec x\tan x+{\sec }^{2}x\right)dx\right)$
$=e^{\sec x}f\left(x\right)+C$
Diff. both sides w.r.t. 'x'
$e^{\sec x}\left(\sec x\tan xf\left(x\right)+\left(\sec x\tan x+{\sec }^{2}x\right)\right)$
$=e^{\sec x}.\sec x\tan xf\left(x\right)+e^{\sec x}{f}^{\prime }\left(x\right)$
$f^{\prime }\left(x\right)={\sec }^{2}x+\tan x\sec x$
$\Rightarrow f\left(x\right)=\tan x+\sec x+c$