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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
Options:
where c is constant of integration, then $f(x)$ is
equal to
Solution:
1512 Upvotes
Verified Answer
The correct answer is:
$x-\sec x$
We have. $\int e^{\operatorname{cin} x}\left(\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-\sec x \tan x) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-1+1-\sec x \tan x) d x$
$=e^{\sin x} f(x)+c$
$\left.\int[ e^{\sin x} \cos x(x-\sec x)+e^{\sin x}(1-\sec x \tan x)\right] d x$
$=e^{\sin x} f(x)+c$
$\int \frac{d}{d x}\left\{e^{\sin x}(x-\sec x)\right\} d x=e^{\sin x} f(x)+c$
$e^{\operatorname{cin} x}(x-\sec x)=e^{\sin x} f(x)+c$
$f(x)=x-\sec x$
$\int e^{\operatorname{sin} x}(x \cos x-\sec x \tan x) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-1+1-\sec x \tan x) d x$
$=e^{\sin x} f(x)+c$
$\left.\int[ e^{\sin x} \cos x(x-\sec x)+e^{\sin x}(1-\sec x \tan x)\right] d x$
$=e^{\sin x} f(x)+c$
$\int \frac{d}{d x}\left\{e^{\sin x}(x-\sec x)\right\} d x=e^{\sin x} f(x)+c$
$e^{\operatorname{cin} x}(x-\sec x)=e^{\sin x} f(x)+c$
$f(x)=x-\sec x$
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