Question

If $\int \sin xd\left(\sec x\right)=f\left(x\right)-g\left(x\right)+c$, then

• A. $f\left(x\right)=\sec x$
• B. $f\left(x\right)=\tan x$
• C. $g\left(x\right)=2x$
• D. $g=-x$

Answer 1

The correct option is B $f\left(x\right)=\tan x$

$\int \sin xd\left(\sec x\right)=\int \sin x\sec x\tan xdx$

$=\int {\tan }^{2}xdx$

$=\int ({\sec }^{2}x-1)dx$

$=\tan x-x+c$

$⟹f\left(x\right)=\tan x,g\left(x\right)=x$