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int(3+2cosx)/((2+3cosx)^2)dxi se q u a l...

`int(3+2cosx)/((2+3cosx)^2)dxi se q u a lto` `((sinx)/(3cosx+2))+c` (b) `((2cosx)/(3sinx+2))+c` `((2cosx)/(3cosx+2))+c` (d) `((2sinx)/(3sinx+2))+c`

A

`((sinx)/(3cosx+2))+c`

B

`((2cosx)/(3sinx+2))+c`

C

`((2cosx)/(3cosx+2))+c`

D

`((2sinx)/(3sinx+2))+c`

Text Solution

Verified by Experts

The correct Answer is:
A
`"Let " I=int(3+2cosx)/((2+3cosx)^(2))dx. `
`"Multiplying " N^(r) " and " D^(r) " by " "cosec"^(2)x, " we get "`
`I=int((3"cosec"^(2)x+2cotx"cosec"x))/((2"cosec"x+3cotx)^(2))dx `
`=-int(-3"cosec"^(2)x-2cotx"cosec"x)/((2"cosec"x+3cotx)^(2))dx `
`=(1)/(2"cosec"x+3cotx)+C=((sinx)/(2+3cosx))+C`
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