Integrating factor of differential equat...

Integrating factor of differential equation `cosx(dy)/(dx)+ysinx=1` is (a) `( b ) (c)cosx (d)` (e) (b) `( f ) (g)tanx (h)` (i) (c) `( d ) (e)secx (f)` (g) (d) `( h ) (i)sinx (j)` (k)

`cosx`

`tanx`

`secx`

`sinx`

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Verified by Experts

The correct Answer is:

`cosx(dy)/(dx)+ysinx=1`
or `(dy)/(dx)+y(sinx)/(cosx)=secx`
`therefore intPdx=int(sinx)/(cosx)dx`
`=-logcosx`
`=log secx`
`therefore` I.F. `=e^(intPdx)=e^(log secx)=secx`

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