Question

If $\int e^{\sin x}\left[\frac{x \cos^{3}x -\sin x}{\cos^{2} x}\right] dx = e^{\sin x}f\left(x\right)+c $ , where c is constant of integration, then $f(x) =$

• A. $\sec \, x - x$
• B. $x - \sec \, x $
• C. $\tan \, x - x$
• D. $x - \tan \, x $

Answer 1

Answer: (B)

We have,
$ \int e^{\sin x}\left(\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right) d x=e^{\sin x} f(x)+c$
$\Rightarrow \int e^{\sin x}(x \cos x-\sec x \tan x) d x=e^{\sin x} f(x)+c$
$\Rightarrow \int e^{\sin x}(x \cos x-1+1-\sec x \tan x) d x $
$= \int^{\sin x} f(x)+c $
$ \Rightarrow \int\left[e^{\sin x} \cos x(x-\sec x)+e^{\sin x}(1-\sec x \tan x)\right] d x $
$=e^{\sin x} f(x)+c $
$ \Rightarrow \int \frac{d}{d x}\left\{e^{\sin x}(x-\sec x)\right\} d x=e^{\sin x} f(x)+c $
$\Rightarrow e^{\sin x}(x-\sec x)=e^{\sin x} f(x)+c $
$ \Rightarrow f(x)=x-\sec x $