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