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Question

Integrating factor of differential equation cosxdydx+ysinx=1 is

A
|cosx|
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B
|tanx|
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C
|secx|
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D
sinx
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Solution

The correct option is C |secx|
cosxdydx+ysinx=1
dydx+ysinxcosx=secx
I.F=ePdx
Now, Pdx=sinxcosxdx=ln|cosx|
So. I.F=eln|secx|=|secx|

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