Q48 / 155 · WBJEE 2018 · Mathematics
If $\int e^{\sin x} \cdot\left[\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right] d x=e^{\sin x} f(x)+c$
where c is constant of integration, then $f(x)$ is
equal to
- A$\sec x-x$
- Correct answerB$x-\sec x$
- C$\tan x-x$
- D$x-\tan x$
Solution
We have. $\int e^{\operatorname{cin} x}\left(\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-\sec x \tan x) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-1+1-\sec x \tan x) d x$
$=e^{\sin x} f(x)+c$
$\left.\int[ e^{\sin x} \cos x(x-\sec x)+e^{\sin x}(1-\sec x \tan x)\right] d x$
$=e^{\sin x} f(x)+c$
$\int \frac{d}{d x}\left\{e^{\sin x}(x-\sec x)\right\} d x=e^{\sin x} f(x)+c$
$e^{\operatorname{cin} x}(x-\sec x)=e^{\sin x} f(x)+c$
$f(x)=x-\sec x$
Hence, the correct answer is option: 'B'.