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Maths-

General

Easy

Question

Integrating factor of differential equation $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

$\cos x$
$\tan x$
$\sec x$
$\sin x$

The correct answer is: $sec invisible function application x$

$fraction numerator d y over denominator d x end fraction plus y tan invisible function application x equals sec invisible function application x$
I.F. $equals e to the power of not stretchy integral tan invisible function application x d x end exponent equals e to the power of logsec invisible function application x end exponent equals sec invisible function application x$

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